{"id":4503,"date":"2023-12-19T11:51:36","date_gmt":"2023-12-19T06:21:36","guid":{"rendered":"https:\/\/envisionmathanswerkey.com\/?p=4503"},"modified":"2023-12-20T09:39:33","modified_gmt":"2023-12-20T04:09:33","slug":"envision-math-common-core-grade-8-answer-key-topic-7","status":"publish","type":"post","link":"https:\/\/envisionmathanswerkey.com\/envision-math-common-core-grade-8-answer-key-topic-7\/","title":{"rendered":"enVision Math Common Core Grade 8 Answer Key Topic 7 Understand And Apply The Pythagorean Theorem"},"content":{"rendered":"

Go through the\u00a0enVision Math Common Core Grade 8 Answer Key<\/a>\u00a0Topic 7 Understand And Apply The Pythagorean Theorem<\/strong>\u00a0regularly and improve your accuracy in solving questions.<\/p>\n

enVision Math Common Core 8th Grade Answers Key Topic 7 Understand And Apply The Pythagorean Theorem<\/h2>\n

Topic Essential Question<\/strong>
\nHow can you use the Pythagorean Theorem to solve problems?
\nAnswer:
\nThe\u00a0Pythagorean Theorem\u00a0is\u00a0used\u00a0to calculate the steepness of slopes of hills or mountains. A surveyor looks through a telescope toward a measuring stick a fixed distance away, so that the telescope’s line of sight and the measuring stick form a right angle.<\/p>\n

3-ACT MATH OOO<\/strong><\/p>\n

\"Envision
\nGo with the Flow
\nYou may have noticed that when you double the base and the height of a triangle, the area is more than doubled. The same is true for doubling the sides of a square or the radius of a circle. So what is the relationship? Think about this during the 3-Act Mathematical Modeling lesson.
\n\"Envision<\/p>\n

Topic 7 enVision STEM Project<\/h3>\n

Did You Know?
\nOver two billion people will face water shortages by 2050 according to a 2015 United Nations Environment Program report.
\n\"Envision
\nRainwater can be collected and stored for use in irrigation, industrial uses, flushing toilets, washing clothes and cars, or it can be purified for use as everyday drinking water.
\nThis alternative water source reduces the use of fresh water from reservoirs and wells.
\n\"Envision
\nUsing water wisely saves money on water and energy bills and extends the life of supply and wastewater facilities.
\n\"Envision
\nRoofs of buildings or large tarps are used to collect rainwater.
\nA rainwater collection system for a building roof that measures 28 feet by 40 feet can provide 700 gallons of water-enough water to support two people for a year\u2014from a rainfall of 1.0 inch.
\n\"Envision
\nEven a 5 foot by 7-foot tarp can collect 2 gallons of water from a rainfall total of only 0.1 in.<\/p>\n

The rainwater harvesting market is expected to grow 5% from 2016 to 2020.
\n\"Envision<\/p>\n

Your Task: Rainy Days
\n\"Envision
\nRainwater collection is an inexpensive way to save water in areas where it is scarce. One inch of rain falling on a square roof with an area of 100 ft\u00b2 collects 62 gallons of water that weighs over 500 pounds. You and your classmates will research the necessary components of a rainwater collection system. Then you will use what you know about right triangles to design a slanted roof system that will be used to collect rainwater.
\nAnswer:
\nIt is given that
\nRainwater collection is an inexpensive way to save water in areas where it is scarce. One inch of rain falling on a square roof with an area of 100 ft\u00b2 collects 62 gallons of water that weighs over 500 pounds
\nNow,
\nThe necessary components of a rainwater collection system are:
\nA) Catchments B) Coarse mesh C) Gutters D) Conduits E) First-flushing F) Filter G) Storage facility H) Recharge Structures<\/p>\n

\"Understand<\/p>\n

Topic 7 GET READY!<\/h3>\n

Review What You Know!<\/strong><\/p>\n

Vocabulary
\nChoose the best term from the box to complete each definition.
\ncube root
\ndiagonal
\nisosceles triangle
\nperimeter
\nright triangle
\nsquare root<\/p>\n

Question 1.
\nThe __________ of a number is a factor that when multiplied by itself gives the number.
\nAnswer:
\nWe know that,
\nThe “Square root” of a number is a factor that when multiplied by itself gives the number
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is a “Square root”<\/p>\n

Question 2.
\nA _________ is a line segment that connects two vertices of a polygon and is not the side.
\nAnswer:
\nWe know that,
\nA “Diagonal” is a line segment that connects two vertices of a polygon and is not the side
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is a “Diagonal”<\/p>\n

Question 3.
\nThe _________ of a figure is the distance around it.
\nAnswer:
\nWe know that,
\nThe “Perimeter” of a figure is the distance around it
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is the “Perimeter”<\/p>\n

Question 4.
\nA ___________ is a triangle with one right angle.
\nAnswer:
\nWe know that,
\nA “Right triangle” is a triangle with one right angle
\nHence, from the above,
\nWe can conclude that the best term to complete the given definition is a “Right angle”<\/p>\n

Simplify Expressions with Exponents<\/strong><\/p>\n

Simplify the expression.
\nQuestion 5.
\n32<\/sup> + 42<\/sup>
\nAnswer:
\nThe given expression is: 32<\/sup> + 42<\/sup>
\nSo,
\n32<\/sup> + 42<\/sup>
\n= (3 \u00d7 3) + (4 \u00d7 4)
\n= 9 + 16
\n= 25<\/p>\n

Question 6.
\n22<\/sup> + 52<\/sup>
\nAnswer:
\nThe given expression is: 22<\/sup> + 52<\/sup>
\nSo,
\n2\u00b2 + 52<\/sup>
\n= (2 \u00d7 2) + (5 \u00d7 5)
\n= 4 + 25
\n= 29<\/p>\n

Question 7.
\n102<\/sup> – 82<\/sup>
\nAnswer:
\nThe given expression is: 102<\/sup> – 82<\/sup>
\nSo,
\n102<\/sup> – 82<\/sup>
\n= (10 \u00d7 10) – (8 \u00d7 8)
\n= 100 – 64
\n= 36<\/p>\n

Square Roots<\/strong><\/p>\n

Determine the square root.
\nQuestion 8.
\n\\(\\sqrt {81}\\)
\nAnswer:
\nThe given expression is: \\(\\sqrt{81}\\)
\nHence,
\n\\(\\sqrt{81}\\) = 9<\/p>\n

Question 9.
\n\\(\\sqrt {144}\\)
\nAnswer:
\nThe given expression is: \\(\\sqrt{144}\\)
\nHence,
\n\\(\\sqrt{144}\\) = 12<\/p>\n

Question 10.
\n\\(\\sqrt {225}\\)
\nAnswer:
\nThe given expression is: \\(\\sqrt{225}\\)
\nHence,
\n\\(\\sqrt{225}\\) = 15<\/p>\n

Distance on a Coordinate Plane<\/strong><\/p>\n

Determine the distance between the two points.
\nQuestion 11.
\n\"Envision
\nAnswer:
\nThe given graph is:
\n\"Envision
\nFrom the given graph,
\nThe given points are: (2, 5), (7, 5)
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nWe know that,
\nDistance between 2 points = \u221a(x2<\/sub> – x1<\/sub>)2<\/sup> + (y2<\/sub> – y1<\/sub>)2<\/sup>
\n= \u221a(7 – 2)2<\/sup> + (5 – 5)2<\/sup>
\n= \\(\\sqrt{5\u00b2}\\)
\n= 5 units
\nHence, from the above,
\nWe can conclude that the distance between the given points is: 5 units<\/p>\n

Question 12.
\n\"Envision
\nAnswer:
\nThe given graph is:
\n\"Envision
\nFrom the given graph,
\nThe given points are: (3, 2), (3, 9)
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nWe know that,
\nDistance between 2 points =\u221a(x2<\/sub> – x1<\/sub>)2<\/sup> + (y2<\/sub> – y1<\/sub>)2<\/sup>
\n= \u221a(3 – 3)2<\/sup> + (9 – 2)2<\/sup>
\n= \\(\\sqrt{7\u00b2}\\)
\n= 7 units
\nHence, from the above,
\nWe can conclude that the distance between the given points is: 7 units<\/p>\n

\"Understand<\/p>\n

Language Development<\/strong><\/p>\n

Complete the word map using key terms, examples, or illustrations related to the Pythagorean Theorem and its Converse.
\n\"Envision
\n\"Envision
\nAnswer:
\n\"\"
\n\"\"<\/p>\n

Topic 7 PICK A PROJECT<\/h3>\n

PROJECT 7A
\nWhere would you like to bike ride in your neighborhood?
\nPROJECT: PLAN A METRIC CENTURY RIDE
\n\"Envision<\/p>\n

PROJECT 7B
\nWhat designs have you seen on kites?
\nPROJECT: BUILD A KITE
\n\"Envision<\/p>\n

PROJECT 7C
\nWhat buildings in your community have unusual shapes as part of their structure or design?
\nPROJECT: MAKE A SCRAPBOOK
\n\"Envision<\/p>\n

PROJECT 7D
\nWhat geometric designs have you noticed on your clothes?
\nPROJECT: DESIGN A FABRIC TEMPLATE
\n\"Envision<\/p>\n

3-ACT MATH<\/strong><\/p>\n

3-Act Mathematical Modeling: Go with the Flow<\/p>\n

\"Envision<\/p>\n

\u0410\u0421\u0422 1<\/strong>
\nQuestion 1.
\nAfter watching the video, what is the first question that comes to mind?
\nAnswer:<\/p>\n

\"Understand<\/p>\n

Question 2.
\nWrite the Main Question you will answer.
\nAnswer:<\/p>\n

Question 3.
\nMake a prediction to answer this Main Question.
\n_________ % will fit in the third square.
\nAnswer:<\/p>\n

Question 4.
\nConstruct Arguments Explain how you arrived at your prediction.
\n\"Envision
\nAnswer:<\/p>\n

ACT 2<\/strong>
\nQuestion 5.
\nWhat information in this situation would be helpful to know? How would you use that information?
\nAnswer:<\/p>\n

Question 6.
\nUse Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
\nAnswer:<\/p>\n

Question 7.
\nModel With Math
\nRepresent the situation using mathematics. Use your representation to answer the Main Question.
\nAnswer:<\/p>\n

Question 8.
\nWhat is your answer to the Main Question? Does it differ from your prediction? Explain.
\n\"Envision
\nAnswer:<\/p>\n

\u0410\u0421\u0422 3<\/strong>
\nQuestion 9.
\nWrite the answer you saw in the video.
\nAnswer:<\/p>\n

Question 10.
\nReasoning<\/strong> Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
\nAnswer:<\/p>\n

Question 11.
\nMake Sense and Persevere Would you change your model now that you know the answer? Explain.
\n\"Envision
\nAnswer:<\/p>\n

Act 3
\nExtension
\nReflect
\nQuestion 12.
\nModel with Math<\/strong>
\nExplain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?
\n\"Envision
\nAnswer:<\/p>\n

Question 13.
\nReason Abstractly<\/strong> How did you represent the situation using symbols? How did you use those symbols to solve the problem?
\n\"Envision
\nAnswer:<\/p>\n

SEQUEL<\/strong>
\nQuestion 14.
\nConstruct Arguments Explain why you can use an area formula when the problem involves comparing volumes.
\n\"Envision
\nAnswer:<\/p>\n

Lesson 7.1 Understand the Pythagorean Theorem<\/h3>\n

Explain It!<\/strong>
\nKelly drew a right triangle on graph paper. Kelly says that the sum of the areas of squares with side lengths a and b is the same as the area of a square with side length c.
\n\"Envision<\/p>\n

I can… use the Pythagorean Theorem to find unknown sides of triangles.<\/p>\n

A. Do you agree with Kelly? Explain.
\nAnswer:
\nIt is given that
\nKelly drew a right triangle on graph paper. Kelly says that the sum of the areas of squares with side lengths a and b is the same as the area of a square with side length c.
\nWe know that,
\nAccording to the Pythagorean theorem,
\nHypotenuse\u00b2 = Side length 1\u00b2 + Side length 2\u00b2
\nSo,
\nFrom the given graph,
\nWe can observe that
\nSide length 1 is: a
\nSide length 2 is: b
\nThe hypotenuse is: c
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that we can agree with Kelly<\/p>\n

B. Sam drew a different right triangle with side lengths a = 5, b = 12, and c = 13. Is the relationship Kelly described true for Sam’s right triangle? Explain.
\nAnswer:
\nIt is given that
\nSam drew a different right triangle with side lengths a = 5, b = 12, and c = 13
\nNow,
\nFrom part (a),
\nThe relation according to Kelly is:
\nc\u00b2 = a\u00b2 + b\u00b2
\nSo,
\n13\u00b2 = 12\u00b2 + 5\u00b2
\n169 = 144 + 25
\n169 = 169
\nHence, from the above,
\nWe can conclude that the relationship Kelly described is true for Sam’s right-angled triangle<\/p>\n

Focus on math practices
\nGeneralize Kelly draws another right triangle. What would you expect to be the relationship between the areas of the squares drawn on each side of the triangle? Explain.
\nAnswer:
\nIt is given that Kelly draws another right triangle
\nHence,
\nIf in a triangle, the\u00a0square\u00a0on one of the\u00a0sides\u00a0equals the sum of the\u00a0squares\u00a0on the remaining two\u00a0sides of the triangle, then the angle contained by the remaining two\u00a0sides of the triangle\u00a0is right.”<\/p>\n

\"Understand<\/p>\n

Essential Question<\/strong>
\nHow does the Pythagorean Theorem relate to the side lengths of a right triangle?
\nAnswer:
\nThe Pythagorean equation relates the sides of a right triangle in a simple way so that if the\u00a0lengths\u00a0of any two\u00a0sides are\u00a0known the\u00a0length\u00a0of the third\u00a0side can\u00a0be found. Another corollary of the\u00a0theorem\u00a0is that in any\u00a0right triangle, the hypotenuse is greater than any one of the other\u00a0sides\u00a0but less than their sum.<\/p>\n

Try It!<\/strong><\/p>\n

A right triangle has side lengths 15 centimeters, 25 centimeters, and 20 centimeters. How can you use the Pythagorean Theorem to write an equation that describes how the side lengths are related?
\na2<\/sup> + b2<\/sup> = c2<\/sup>
\n_______2<\/sup> + ________2<\/sup> = _________2<\/sup>
\n_________ + _________ = ________
\nAnswer:
\nIt is given that
\nA right triangle has side lengths 15 centimeters, 25 centimeters, and 20 centimeters.
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nThe hypotenuse has the greatest length in the right triangle
\nNow,
\nLet the hypotenuse be c
\nLet the other two sides be a and b
\nSo,
\nFrom the given information,
\nc = 25 centimeters, a = 15 centimeters,and b = 20 centimeters
\nSo,
\nAccording to the Pythagorean Theorem,
\n25\u00b2 = 15\u00b2 + 20\u00b2
\n625 = 225 + 400
\n625 = 625
\nHence, from the above,
\nWe can conclude that we proved how the Pythagorean Theorem relates to the lengths of the right triangle<\/p>\n

Convince Me!
\nHow do you know that the geometric proof of the Pythagorean Theorem shown above can be applied to all right triangles?
\nAnswer:
\nIt\u00a0can\u00a0be proven using the law of cosines or as follows: Let ABC be a\u00a0triangle\u00a0with side lengths a, b, and c, with a2<\/sup>\u00a0+ b2<\/sup>\u00a0= c2<\/sup>.\u00a0 Therefore, the angle between the side of lengths a and b in the original triangle is a right angle. This proof of the converse makes use of the Pythagorean theorem\u00a0itself.<\/p>\n

Try It!<\/strong><\/p>\n

A right triangle has a hypotenuse length of 32 meters. It has one leg with a length of 18 meters. What is the length of the other leg? Express your answer as a square root.
\nAnswer:
\nIt is given that
\nA right triangle has a hypotenuse length of 32 meters. It has one leg with a length of 18 meters.
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the legs
\nNow,
\nLet the other leg be x
\nSo,
\n32\u00b2 = 18\u00b2 + x\u00b2
\nx\u00b2 = 32\u00b2 – 18\u00b2
\nx\u00b2 = 1024 – 324
\nx\u00b2 = 700
\nx = \\(\\sqrt{700}\\)
\nHence, from the above,
\nWe can conclude that the length of the other leg is: \\(\\sqrt{700}\\)<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

The Pythagorean Theorem is an equation that relates the side lengths of a right triangle, a2<\/sup> + b2<\/sup> = c2<\/sup>, where a and b are the legs of a right triangle and c is the hypotenuse.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How does the Pythagorean Theorem relate to the side lengths of a right triangle?
\nAnswer:
\nThe Pythagorean equation relates the sides of a right triangle in a simple way so that if the\u00a0lengths\u00a0of any two\u00a0sides are\u00a0known the\u00a0length\u00a0of the third\u00a0side can\u00a0be found. Another corollary of the\u00a0theorem\u00a0is that in any\u00a0right triangle, the hypotenuse is greater than any one of the other\u00a0sides\u00a0but less than their sum.<\/p>\n

Question 2.
\nUse Structure<\/strong> A side of each of the three squares forms a side of a right triangle.
\nWould any three squares form the sides of a right triangle? Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\nA side of each of the three squares forms a side of a right triangle.
\nNow,
\nWe know that,
\nThe length of all the sides in a square is equal
\nNow,
\nFrom the given figure,
\nWe can observe that
\nEach side of a square from the three squares form a right triangle
\nHence, from the above,
\nWe can conclude that the three squares form the sides of a right triangle<\/p>\n

Question 3.
\nConstruct Arguments<\/strong> Xavier said the missing length is about 18.5 units. Without calculating, how can you tell that Xavier solved incorrectly?
\n\"Envision
\nAnswer:
\nIt is given that
\nXavier said the missing length is about 18.5 units
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nThe length of the hypotenuse is the greatest
\nNow,
\nThe given right triangle is:
\n\"Envision
\nSo,
\nAccording to the Pythagorean Theorem,
\nThe length of the missing side should be greater than 21 and 26
\nHence, from the above,
\nWe can conclude that Xavier calculated incorrectly<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nA right triangle has leg lengths of 4 inches and 5 inches. What is the length of the hypotenuse? Write the answer as a square root and round to the nearest tenth of an inch.
\nAnswer:
\nIt is given that
\nA right triangle has leg lengths of 4 inches and 5 inches
\nNow,
\nWe know that,
\nAccording to the Pythagorean theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nc\u00b2 = 4\u00b2 + 5\u00b2
\nc\u00b2 = 16 + 25
\nc\u00b2 = 41
\nc = \\(\\sqrt{41}\\)
\nc = 6.4 inches
\nHence, from the above,
\nWe can conclude that the length of the hypotenuse is: 6.4 inches<\/p>\n

Question 5.
\nFind the missing side length to the nearest tenth of afoot.
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\n14\u00b2 = 8\u00b2 + b\u00b2
\nb\u00b2= 14\u00b2 – 8\u00b2
\nb\u00b2 = 196 – 64
\nb\u00b2 = 132
\nb = \\(\\sqrt{132}\\)
\nb = 11.5 feet
\nHence, from the above,
\nWe can conclude that the length of the missing side is: 11.5 feet<\/p>\n

Question 6.
\nFind the missing side length to the nearest tenth of a millimeter.
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nc\u00b2 = (3.7)\u00b2 + (7.5)\u00b2
\nc\u00b2 = 13.69 + 56.25
\nc\u00b2 = 69.94
\nc = \\(\\sqrt{69.94}\\)
\nc = 8.4 mm
\nHence, from the above,
\nWe can conclude that the length of the missing side is: 8.4 mm<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Leveled Practice In 7 and 8, find the missing side length of each triangle.
\nQuestion 7.
\n\"Envision
\nThe length of the hypotenuse is ________ units.
\nAnswer:
\n\"\"<\/p>\n

Question 8.
\n\"Envision
\nThe length of leg b is about ________ inches.
\nAnswer:
\n\"\"<\/p>\n

Question 9.
\nWhat is the length of the hypotenuse of the triangle when x = 15? Round your answer to the nearest tenth of a unit.
\n\"Envision
\nAnswer:
\nThe given right angle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nc\u00b2 = (3x)\u00b2 + (4x + 4)\u00b2
\nNow,
\nWhen x = 15,
\nc\u00b2 = (3 \u00d7 15)\u00b2 + (4 \u00d7 15 + 4)\u00b2
\nc\u00b2 = 45\u00b2 + 64\u00b2
\nc\u00b2 = 2,025 + 4,096
\nc\u00b2 = 6,121
\nc = \\(\\sqrt{6,121}\\)
\nc = 78.2 units
\nHence, from the above,
\nWe can conclude that the length of the hypotenuse when x= 15 is: 78.2 units<\/p>\n

Question 10.
\nWhat is the length of the missing side rounded to the nearest tenth of a centimeter?
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\na\u00b2 = (12.9)\u00b2 + (15.3)\u00b2
\na\u00b2 = 166.41 + 234.09
\na\u00b2 = 400.5
\na = \\(\\sqrt{400.5}\\)
\na = 20 cm
\nHence, from the above,
\nWe can conclude that the length of the side a is: 20 cm<\/p>\n

Question 11.
\nUse the Pythagorean Theorem to find the unknown side length of the right triangle.
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nc\u00b2 = (10)\u00b2 + (24)\u00b2
\nc\u00b2 = 100 + 576
\nc\u00b2 = 676
\nc = \\(\\sqrt{676}\\)
\nc = 26 m
\nHence, from the above,
\nWe can conclude that the length of the side a is: 26 m<\/p>\n

Question 12.
\nWhat is the length of the unknown leg of the right triangle rounded to the nearest tenth of afoot?
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\n9\u00b2 = 2\u00b2 + b\u00b2
\nb\u00b2 = 9\u00b2 – 2\u00b2
\nb\u00b2 = 81 – 4
\nb\u00b2 = 77
\nb = \\(\\sqrt{77}\\)
\nb = 8.8 foot
\nHence, from the above,
\nWe can conclude that the length of the unknown leg is: 8.8 foot<\/p>\n

Question 13.
\nA student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 26 centimeters. The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
\na. Find the length of the hypotenuse of the right triangle to the nearest tenth of a centimeter.
\nAnswer:
\nIt is given that
\nA student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the length of the other leg is 26 centimeters. The student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nc\u00b2 = 32\u00b2 + 26\u00b2
\nc\u00b2 = 1,024 + 676
\nc\u00b2 = 1,700
\nc = \\(\\sqrt{1,700}\\)
\nc = 41.2 centimeters
\nHence, from the above,
\nWe can conclude that the length of the hypotenuse of a right triangle is: 41.2 centimeters<\/p>\n

b. What mistake might the student have made?
\nAnswer:
\nIt is given that
\nThe student incorrectly says that the length of the hypotenuse is 7.6 centimeters.
\nBut,
\nFrom part (a),
\nThe length of the hypotenuse is: 41.2 centimeters
\nHence, from the above,
\nWe can conclude that the mistake the student might make is the misinterpretation of the length of the hypotenuse<\/p>\n

Question 14.
\nFind the length of the unknown leg of the right triangle.
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\n(37.25)\u00b2 = (12.25)\u00b2 + b\u00b2
\nb\u00b2 = (37.25)\u00b2 – (12.25)\u00b2
\nb\u00b2 = 1,387.56 – 150.06
\nb\u00b2 = 1,237.5
\nb = \\(\\sqrt{1,237.5}\\)
\nb = 35.17 units
\nHence, from the above,
\nWe can conclude that the length of the unknown leg is: 35.17 units<\/p>\n

Question 15.
\nHigher-Order Thinking<\/strong> A right triangle has side lengths 12 centimeters and 14 centimeters. Name two possible side lengths for the third side, and explain how you solved for each.
\nAnswer:
\nIt is given that
\nA right triangle has side lengths of
\n12 centimeters and 14 centimeters.
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nNow,
\nLet the length of the third side be x
\nSo,
\nThe possible lengths of the third side may be:
\nx < 12 centimeters and x > 14 centimeters
\nHence, from the above,
\nWe can conclude that the two possible side lengths for the third side are:
\nx < 12 centimeters and x > 14 centimeters<\/p>\n

Assessment Practice<\/strong>
\nQuestion 16.
\nWhich right triangle has a hypotenuse that is about 39 feet long?
\n\"Envision
\nAnswer:
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nFor Options A and D:
\nA) c\u00b2 = 30\u00b2 + 15\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D) c\u00b2 = 30\u00b2 + 14\u00b2
\n= 33.54 ft\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 33.10 ft
\nFor Options B and C:
\nB) c\u00b2 = 36\u00b2 + 12\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C) c\u00b2 = 36\u00b2 + 15\u00b2
\n= 37.94 ft\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 39 ft
\nHence, from the above,
\nWe can conclude that Option C matches the given situation<\/p>\n

Question 17.
\nWhich right triangle does NOT have an unknown leg length of about 33 cm?
\n\"Envision
\nAnswer:
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the hypotenuse
\na and b are the lengths of the legs
\nSo,
\nFor Options A and D:
\nA) b\u00b2 = 35\u00b2 – 11\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D) b\u00b2 = 34\u00b2 – 9\u00b2
\n= 33.22 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 32.78 cm
\nFor Options B and C:
\nB) b\u00b2 = 30\u00b2 – 10\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C) b\u00b2 = 35\u00b2 – 12\u00b2
\n= 28.28 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 32.87 cm
\nHence, from the above,
\nWe can conclude that Option B matches the given situation<\/p>\n

Lesson 7.2 Understand the Converse of the Pythagorean Theorem<\/h3>\n

Solve & Discuss It!<\/strong>
\nKayla has some straws that she will use for an art project. She wants to glue three of the straws onto a sheet of paper, without overlapping, to make the outline of a right triangle. Which three straws could Kayla use to make a right triangle? Explain.
\n\"Envision
\nI can… use the Converse of the Pythagorean Theorem to identify right triangles.
\nAnswer:
\nIt is given that
\nKayla has some straws that she will use for an art project. She wants to glue three of the straws onto a sheet of paper, without overlapping, to make the outline of a right triangle.
\nNow,
\nWe know that,
\nThe converse of the Pythagorean Theorem states that if the square of the third side of a triangle is equivalent to the sum of its two shorter sides, then it must be a right triangle i.e.,
\nIf c\u00b2 = a\u00b2+ b\u00b2, then the given triangle is a right triangle
\nSo,
\nFrom the given straws,
\nwe can observe that
\nThe straws numbered 3, 4, and 5 can be glued to make the outline of a right triangle
\nThe straws numbered 5, 12, and 13 can be glued to make the outline of a right triangle
\nHence, from the above,
\nWe can conclude that there are 2 pairs of straws i.e., (3, 4, 5) and (12, 5, 13) to make the outline of a right triangle<\/p>\n

Look for Relationships
\nHow could you use the Pythagorean Theorem to determine whether the lengths form a right triangle?
\nAnswer:
\nAccording to the converse of the Pythagorean Theorem,
\nIf the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.<\/p>\n

Focus on math practices
\nUse Structure Could Kayla use the straws that form a right triangle to make a triangle that is not a right triangle? Explain.
\nAnswer:
\nFrom the given straws,
\nWe can observe that the pair (6, 7, 4) can’t form a right triangle but (3, 4, 5) can form a right triangle
\nHence, from the above,
\nWe can conclude that Kayla can use the straws that form a right triangle to make a triangle that is not a right triangle<\/p>\n

Essential Question<\/strong>
\nHow can you determine whether a triangle is a right triangle?
\nAnswer:
\nWe can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem
\nHence,
\nAccording to the converse of the Pythagorean Theorem,
\nIf the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.<\/p>\n

Try It!<\/strong><\/p>\n

A triangle has side lengths 4 inches, 5 inches, and 7 inches. Is the triangle a right triangle?
\n\"Envision
\nIs a2<\/sup> + b2<\/sup> equal to c2<\/sup>? _______
\nIs the triangle a right triangle? _________
\nAnswer:
\nIt is given that
\nA triangle has side lengths 4 inches, 5 inches, and 7 inches.
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse that has the longest side length
\na and b are the lengths of the legs
\nSo,
\n7\u00b2 = 4\u00b2 + 5\u00b2
\n49 = 16 + 25
\n49 = 41
\nSo,
\n49 \u2260 41
\nHence, from the above,
\nWe can conclude that
\nc\u00b2 \u2260 a\u00b2+ b\u00b2
\nThe given triangle is not a right triangle<\/p>\n

Convince Me!
\nExplain the proof of the Converse of the Pythagorean Theorem in your own words.
\nAnswer:
\nThe\u00a0converse of the Pythagorean Theorem\u00a0is:
\nIf the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.<\/p>\n

Try It!<\/strong><\/p>\n

A triangle has side lengths 10 feet, \\(\\sqrt {205}\\) feet, and \\(\\sqrt {105}\\) feet. Is this a right triangle? Explain.
\nAnswer:
\nIt is given that
\nA triangle has side lengths 10 feet, \\(\\sqrt {205}\\) feet, and \\(\\sqrt {105}\\) feet.
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2, then
\nThe given triangle is a right triangle
\nWe know that,
\nc is the length of the hypotenuse that has the longest side in a right triangle
\nSo,
\n(\\(\\sqrt{205}\\))\u00b2 =(\\(\\sqrt{105}\\))\u00b2 + 10\u00b2
\n205 = 105 + 100
\n205 = 205
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the triangle with the given side lengths is a right triangle<\/p>\n

Try It!<\/strong><\/p>\n

A triangle is inside a trapezoid. Is the triangle a right triangle? Explain.
\n\"Envision
\nAnswer:
\nIt is given that a triangle is inside a trapezoid
\nNow,
\nThe given trapezoid is:
\n\"Envision
\nFrom the given trapezoid,
\nThe sides of the triangle are: 17 in., 15 in., \\(\\sqrt{514}\\) in.
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen, the given triangle is a right triangle
\nSo,
\n(\\(\\sqrt{514}\\))\u00b2 = 17\u00b2 + 15\u00b2
\n514 = 289 + 225
\n514 = 514
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the triangle that is in the trapezoid is a right triangle<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

The Converse of the Pythagorean Theorem states that if the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, the triangle is a right triangle.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How can you determine whether a triangle is a right triangle?
\nAnswer:
\nWe can determine the triangle is a right triangle by using the converse of the Pythagorean Theorem
\nHence,
\nAccording to the converse of the Pythagorean Theorem,
\nIf the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.<\/p>\n

Question 2.
\nConstruct Arguments<\/strong> A triangle has side lengths of 3 centimeters, 5 centimeters, and 4 centimeters. Abe used the Converse of the Pythagorean Theorem to determine whether it is a right triangle.
\n32<\/sup> + 52<\/sup> \\(\\underline{\\underline{?}}\\) 42<\/sup>
\n9 + 25 \\(\\underline{\\underline{?}}\\) 16
\n34 \u2260 16
\nAbe concluded that it is not a right triangle. Is Abe correct? Explain.
\nAnswer:
\nIt is given that
\nA triangle has side lengths of 3 centimeters, 5 centimeters, and 4 centimeters
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is\u00a0 a right triangle
\nWe know that,
\nThe hypotenuse has the length of the longest side
\nSo,
\n5\u00b2= 3\u00b2 + 4\u00b2
\n25 = 9 + 16
\n25 = 25
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the given triangle is a right triangle and Abe is not correct<\/p>\n

Question 3.
\nUse Structure<\/strong> When you are given three side lengths for a triangle, how do you know which length to substitute for a, b, or c in the Pythagorean Theorem?
\nAnswer:
\nFirst, we will find the squares of all lengths, then we will check which two squares of sides are equal to the square of the third side as per Pythagoras theorem. Hence the two sides would according be a and b and the third side will become c<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nIs the triangle a right triangle? Explain.
\n\"Envision
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf c\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo,
\n10\u00b2 = 6\u00b2 + 8\u00b2
\n100 = 36 + 64
\n100 = 100
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the given triangle is a right triangle<\/p>\n

Question 5.
\nIs the triangle a right triangle? Explain.
\n\"Envision
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf c\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo,
\n8\u00b2 = (\\(\\sqrt{26}\\))\u00b2 + (\\(\\sqrt{28}\\))\u00b2
\n64 = 26 + 28
\n64 = 54
\nSo,
\nc\u00b2 \u2260 a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the given triangle is not a right triangle<\/p>\n

Question 6.
\nIs the purple triangle a right triangle? Explain.
\n\"Envision
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf c\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo,
\n(20.8)\u00b2 = 14\u00b2 + (15.5)\u00b2
\n432.64 = 196 + 240.25
\n432.64 = 436.25
\nSo,
\nc\u00b2 \u2260 a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the purple triangle is not a right triangle<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Leveled Practice In 7 and 8, determine whether each triangle is a right triangle.
\nQuestion 7.
\n\"Envision
\nIs the triangle a right triangle? ________
\nAnswer:
\n\"\"
\nHence, from the above,
\nWe can conclude that the given triangle is not a right triangle<\/p>\n

Question 8.
\n\"Envision
\nIs the triangle a right triangle? _________
\nAnswer:
\n\"\"
\nHence, from the above,
\nWe can conclude that the given triangle is a right triangle<\/p>\n

Question 9.
\nCan the sides of a right triangle have lengths 5, 15, and \\(\\sqrt {250}\\)? Explain.
\nAnswer:
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nThe hypotenuse has the longest side length
\nSo,
\n(\\(\\sqrt{250}\\))\u00b2 = 5\u00b2 + 15\u00b2
\n250 = 25 + 225
\n250 = 250
\nHence, from the above,
\nWe can conclude that the sides of a right triangle have lengths 5, 15, and \\(\\sqrt {250}\\)<\/p>\n

Question 10.
\nIs \u2206PQR a right triangle? Explain.
\n\"Envision
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf c\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo,
\n(6.25)\u00b2 = (3.75)\u00b2 + 5\u00b2
\n39.0625 = 14.0625 + 25
\n39.0625 = 39.0625
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that \u0394PQR is a right triangle<\/p>\n

Question 11.
\nThe green triangle is set inside a rectangle. Is the green triangle a right triangle? Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\nThe green triangle is set inside a rectangle
\nNow,
\nThe given figure is:
\n\"Envision
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf c\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo,
\n21\u00b2 = (\\(\\sqrt{282}\\))\u00b2 + (\\(\\sqrt{159}\\))\u00b2
\n441 = 282 + 159
\n441 = 441
\nSo,
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that the given green triangle is a right triangle<\/p>\n

Question 12.
\nThe side lengths of three triangles are shown. Which of the triangles are right triangles?
\n\"Envision
\nAnswer:
\nIt is given that
\nThe side lengths of the three triangles are shown in the table
\nNow,
\nThe given table is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf c\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo
\nFor Triangle 1,
\n(\\(\\frac{5}{7}\\))\u00b2 = (\\(\\frac{3}{7}\\))\u00b2 + (\\(\\frac{4}{7}\\))\u00b2
\n25 = 16 + 9
\n25 = 25
\nFor Triangle 2,
\n15\u00b2 = 8\u00b2 + 8\u00b2
\n225 = 64 + 64
\n225 = 128
\nFor Triangle 3,
\n(\\(\\frac{13}{17}\\))\u00b2 = (\\(\\frac{12}{17}\\))\u00b2 + (\\(\\frac{5}{17}\\))\u00b2
\n169 = 144 + 25
\n169 = 169
\nSo,
\nThe condition
\nc\u00b2 = a\u00b2 + b\u00b2
\nis true for the Triangles 1 and 3
\nHence, from the above,
\nWe can conclude that Triangle 1 and Triangle 3 is a right triangle<\/p>\n

Question 13.
\nConstruct Arguments<\/strong> Three students draw triangles with the side lengths shown. All three say that their triangle is a right triangle. Which students are incorrect? What mistake might they have made?
\nStudent 1: 22, 33, 55
\nStudent 2: 44, 33, 77
\nStudent 3: 33, 44, 55
\nAnswer:
\nIt is given that
\nThree students draw triangles with the side lengths shown. All three say that their triangle is a right triangle.
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nNow,
\nFor student 1,
\n55\u00b2 = 33\u00b2 + 22\u00b2
\n3,025 = 1,089 + 484
\n3,025 = 1,573
\nFor student 2,
\n77\u00b2= 44\u00b2 + 33\u00b2
\n5,929 = 1,936 + 1,089
\n5,929 = 3,025
\nFor student 3,
\n55\u00b2 = 44\u00b2 + 33\u00b2
\n3,025 = 1,936 + 1,089
\n3,025 = 3,025
\nSo,
\nThe condition
\nc\u00b2 = a\u00b2 + b\u00b2
\nis false for the side lengths of the triangles that are drawn by students 1 and 2
\nHence, from the above,
\nWe can conclude that student 1 and student 2 are incorrect<\/p>\n

Question 14.
\nModel with Math
\n\u2206JKL is an isosceles triangle. Is KM the height of \u2206JKL? Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\n\u0394JKL is an isosceles triangle
\nNow,
\nTo find whether KM is the height of \u0394JKL,
\nFind out whether \u0394KLM is a right triangle or not
\nNow,
\nThe given triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nSo,
\nFrom \u0394KLM,
\n(\\(\\sqrt{340}\\))\u00b2 = 13\u00b2 + 14\u00b2
\n340 = 169 + 196
\n340 = 365
\nSo,
\nc\u00b2 \u2260 a\u00b2 + b\u00b2
\nSo,
\n\u0394KLM is not a right triangle
\nHence, from the above,
\nWe can conclude that KM is not the height of \u0394JKL<\/p>\n

Question 15.
\nHigher-Order Thinking<\/strong> The side lengths of three triangles are given.
\nTriangle 1: \\(\\sqrt{229}\\) units, \\(\\sqrt{225}\\) units, 22 units
\nTriangle 2: \\(\\sqrt{11 \\frac{1}{3}}\\) units, \\(\\sqrt{13 \\frac{2}{3}}\\) units, 5 units
\nTriangle 3: 16 units, 17 units, \\(\\sqrt{545}\\) units
\na. Which lengths represent the side lengths of a right triangle?
\nAnswer:
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2
\nthen, the triangle is a right triangle
\nNow,
\nFor Triangle 1,
\n(\\(\\sqrt{229}\\))\u00b2 = (\\(\\sqrt{225}\\))\u00b2 + 22\u00b2
\n229 = 225 + 484
\n229 = 709
\nFor Triangle 2,
\n(\\(\\sqrt{\\frac{41}{3}}\\))\u00b2 + (\\(\\sqrt{\\frac{34}{3}}\\))\u00b2 = 5\u00b2
\n\\(\\frac{41}{3}\\) + \\(\\frac{34}{3}\\) = 25
\n25 = 25
\nFor Triangle 3,
\n(\\(\\sqrt{545}\\))\u00b2 = 16\u00b2 + 17\u00b2
\n545 = 256 + 289
\n545 = 545
\nSo,
\nThe condition
\nc\u00b2 = a\u00b2+ b\u00b2
\nis true for triangle 2 and triangle 3
\nHence, from the above,
\nWe can conclude that Triangle 2 and Triangle 3 represent the side lengths of a triangle<\/p>\n

b. For any triangles that are not right triangles, use two of the sides to make a right triangle.
\nAnswer:<\/p>\n

Assessment Practice<\/strong>
\nQuestion 16.
\nWhich shaded triangle is a right triangle? Explain.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2+ b\u00b2,
\nthen the triangle is a right triangle
\nNow,
\nFor \u0394ABC,
\n144\u00b2 = 63\u00b2 + (\\(\\sqrt{9}\\))\u00b2
\n20,736 = 3,969 + 9
\n20,736 = 3,978
\nFor \u0394XYZ,
\n(\\(\\sqrt{144}\\))\u00b2 = (\\(\\sqrt{63}\\))\u00b2 + 9\u00b2
\n144 = 63 + 81
\n144 = 144
\nSo,
\nThe condition c\u00b2 = a\u00b2+ b\u00b2 is true for the shaded triangle XYZ
\nHence, from the above,
\nWe can conclude that the shaded triangle \u0394XYZ is a right triangle<\/p>\n

Question 17.
\nWhich triangle is a right triangle?
\n\"Envision
\nA. Triangle I only
\nB. Triangle II only
\nC. Triangle I and Triangle II
\nD. Neither Triangle I nor Triangle II
\nAnswer:
\nThe given triangles are:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen the triangle is a right triangle
\nNow,
\nFor Triangle I,
\n52\u00b2 = 40\u00b2 + 48\u00b2
\n2,704 = 1,600 + 2,304
\n2,704 = 3,904
\nFor Triangle II,
\n65\u00b2 = 60\u00b2 + 25\u00b2
\n4,225 = 3,600 + 625
\n4,225 = 4,225
\nHence, from the above,
\nWe can conclude that Triangle II only is a right triangle<\/p>\n

Topic 7 MID-TOPIC CHECKPOINT<\/h3>\n

Question 1.
\nVocabulary<\/strong> How are the hypotenuse and the legs of a right triangle related? Lesson 7-1
\nAnswer:
\nThe relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right\u00a0angle are called\u00a0legs<\/p>\n

Question 2.
\nGiven that \u2206QPR has side lengths of 12.5 centimeters, 30 centimeters, and 32.5 centimeters, proves \u2206QPR is a right triangle. Lesson 7-2
\nAnswer:
\nIt is given that
\n\u2206QPR has side lengths of 12.5 centimeters, 30 centimeters, and 32.5 centimeters
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nSo,
\n(32.5)\u00b2 = (12.5)\u00b2 + 30\u00b2
\n1,056.25 = 156.25 + 900
\n1,056.25 = 1,056.25
\nSo,
\nThe condition
\nc\u00b2 = a\u00b2 + b\u00b2
\nis true for the given side lengths of a triangle
\nHence, from the above,
\nWe can conclude that \u0394QPR is a right triangle<\/p>\n

Question 3.
\nElla said that if she knows the lengths of just two sides of any triangle, then she can find the length of the third side by using the Pythagorean Theorem. Is Ella correct? Explain. Lesson 7-1
\nAnswer:
\nElla said that if she knows the lengths of just two sides of any triangle, then she can find the length of the third side by using the Pythagorean Theorem.
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the legs
\nNow,
\nIf we know a and b, then we can find c
\nIf we know b and c, then we can find a
\nIf we know a and c, then we can find b
\nHence, from the above,
\nWe can conclude that Ella is correct<\/p>\n

Question 4.
\nFind the unknown side length. Round to the nearest tenth. Lesson 7-1
\n\"Envision
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nFrom the given triangle,
\nWe can observe that it is a right triangle
\nSo,
\nNow,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the legs
\nNow,
\n8\u00b2 = 4\u00b2 + b\u00b2
\n64 = 16 + b\u00b2
\nb\u00b2 = 64 – 16
\nb\u00b2 = 48
\nb = \\(\\sqrt{48}\\)
\nb = 6.9 cm
\nHence, from the above,
\nWe can conclude that the length of the unknown side is: 6.9 cm<\/p>\n

Question 5.
\nThe height of a shed is 6 m. A ladder leans against the shed with its base 4.5 m away, and its top just reaching the roof. What is the length of the ladder? Lesson 7-1
\n\"\"
\nAnswer:
\nIt is given that
\nThe height of a shed is 6 m. A ladder leans against the shed with its base 4.5 m away, and its top just reaching the roof.
\nNow,
\nFrom the given figure,
\nWe can observe that the ladder looks like the hypotenuse of a right triangle for the shed
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the legs
\nIn this situation,
\nc is the length of the ladder
\na is the height of the shed
\nb is the length of the base
\nSo,
\nc\u00b2 = 6\u00b2 + (4.5)\u00b2
\nc\u00b2 = 36 + 20.25
\nc\u00b2 = 56.25
\nc = \\(\\sqrt{56.25}\\)
\nc = 7.5 m
\nHence, from the above,
\nWe can conclude that the length of the ladder is: 7.5 m<\/p>\n

Question 6.
\nSelect all the sets of lengths that could represent the sides of a right triangle. Lesson 7-2
\n\u2610 5 cm, 10 cm, 15 cm
\n\u2610 7 in., 14 in., 25 in.
\n\u2610 13 m, 84 m, 85 m
\n\u2610 5 ft, 11 ft, 12 ft
\n\u2610 6ft, 9 ft, \\(\\sqrt {117}\\) ft
\nAnswer:
\nLet the given options be named as A, B, C, D, and E
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse and has the longest length
\na and b are the legs
\nSo,
\nA)
\n15\u00b2 = 5\u00b2 + 10\u00b2
\n225 = 25 + 100
\n225 \u2260 125
\nB)
\n25\u00b2 = 7\u00b2 + 14\u00b2
\n625 = 49 + 196
\n625 \u2260 245
\nC)
\n85\u00b2 = 84\u00b2 + 13\u00b2
\n7,225 = 7,056 + 169
\n7,225 = 7,225
\nD)
\n12\u00b2 = 11\u00b2 + 5\u00b2
\n144 = 121 + 25
\n144 \u2260 146
\nE)
\n(\\(\\sqrt{117}\\))\u00b2 = 6\u00b2 + 9\u00b2
\n117 = 36 + 81
\n17 = 117
\nHence, from the above,
\nWe can conclude that the side lengths present in options C and E represent the side lengths of a right triangle<\/p>\n

Topic 7 MID-TOPIC PERFORMANCE TASK<\/h3>\n

Javier is standing near a palm tree. He holds an electronic tape measure near his eyes and finds the three distances shown.<\/p>\n

PART A
\nJavier says that he can now use the Pythagorean Theorem to find the height of the tree. Explain. Use vocabulary terms in your explanation.
\nAnswer:
\nIt is given that
\nJavier says that he can now use the Pythagorean Theorem to find the height of the tree
\nNow,
\nAccording to the Pythagorean Theorem,
\nIn a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides\u201c. The sides of this triangle have been named Perpendicular (The height of the tree), Base, and Hypotenuse.<\/p>\n

PART B
\nFind the height of the tree. Round to the nearest tenth. Show your work.
\n\"Envision
\nAnswer:
\nFrom the given figure,
\nWe can observe that
\nTo find the height of the tree, we have to find the perpendicular distances of the two right triangles
\nNow,
\nLet x be the perpendicular height of the first right triangle
\nLet y be the perpendicular height of the second right triangle
\nSo,
\nThe height of the tree = x + y
\nNow,
\nThe side lengths of the first right triangle are: 25 ft, 7 ft, x ft
\nNow,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the side lengths
\nSo,
\n25\u00b2 = 7\u00b2 + x\u00b2
\nx\u00b2 = 625 – 49
\nx\u00b2 = 576
\nx = \\(\\sqrt{576}\\)
\nx = 24 ft
\nNow,
\nThe side lengths of the second right triangle are: 9 ft, 7ft, y ft
\nSo,
\n9\u00b2 = 7\u00b2 + y\u00b2
\ny\u00b2 = 81 – 49
\ny\u00b2 = 32
\ny = \\(\\sqrt{32}\\)
\ny = 5.6 ft
\nSo,
\nThe height of the tree = 24 + 5.6
\n= 29.6 ft
\nHence, from the above,
\nWe can conclude that the height of the tree to the nearest tenth is: 29.6 ft<\/p>\n

PART C
\nJavier moves backward so that his horizontal distance from the palm tree is 3 feet greater. Will the distance from his eyes to the top of the tree also be 3 feet greater? Explain.
\nAnswer:
\nYes it will be greater, he is moving back 3 feet so what you are doing is taking the leg (a\u00b2) and multiplying it by 3. Once you do, you see triangle 1 has double and so did triangle 2. Triangle 2 was originally 5.6 (rounded to 6) then went up to 26.1. Triangle 1 was originally 24 and went up to 74.7 (or 75)
\nStep-by-step explanation:
\nFor Triangle 1 (when multiplied by 3):
\nWe know that,
\na\u00b2 + b\u00b2 = c\u00b2
\n7\u00b2 + b\u00b2 = 27\u00b2
\n49 + b\u00b2 = 729
\nb\u00b2 = 680
\nb\u00a0 = \\(\\sqrt{680}\\)
\nb = 26.1 ft
\nFor Triangle 2:
\nWe know that,
\na\u00b2 + b\u00b2 = c\u00b2
\n7\u00b2 + b\u00b2 =75\u00b2
\nb\u00b2 = 5576
\nb = \\(\\sqrt{5,576}\\)
\nb = 74.7 ft<\/p>\n

PART D
\nCould Javier change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet? Explain.
\nAnswer:
\nYes, Javier can change his horizontal distance from the tree so that the distance from his eyes to the top of the tree is only 20 feet by moving forward 5 ft<\/p>\n

Lesson 7.3 Apply the Pythagorean Theorem to Solve Problems<\/h3>\n

Solve & Discuss It!<\/strong>
\nCarlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has. Which box can Carlos use to ship the umbrella? Explain.
\n\"Envision
\nI can… use the Pythagorean Theorem to solve problems.
\nAnswer:
\nIt is given that
\nCarlos is giving his friend in another state a new umbrella as a gift. He wants to ship the umbrella in a box he already has.
\nNow,
\nFrom the given figure,
\nWe can observe that the umbrella is 37.5 inches long.
\nNow,
\nIf you observe, each box doesn’t have 37 inches wide, however, they are tridimensional figures, which means they have a certain height.
\nSo,
\nThe smaller box is too small to fit the umbrella.
\nBut,
\nThe medium box is perfect because it has 27 inches wide and 27 inches in height, which is enough to fit the umbrella. These are the best dimensions to send the umbrella.
\nHence, from the above,
\nWe can conclude that the box that can be used by Carlos to ship the umbrella has the dimensions 27 in \u00d7 14 in \u00d7 27 in<\/p>\n

Make Sense and Persevere
\nHow will the umbrella fit inside any of the boxes?
\nAnswer:
\nWe know that,
\nThe box is a 3-d figure
\nSo,
\nTo fit the whole umbrella in the box, we have to put it in a diagonal manner i.e., like the hypotenuse of a right triangle<\/p>\n

Focus on math practices
\nConstruct Arguments Tim says that the diagonal of any of the boxes will always be longer than the sides. Is Tim correct? Explain.
\nAnswer:
\nWe know that,
\nIf we consider a square or any 2-d figure or any 3-d figure that consists of 1 right angle,
\nthen, the diagonal will divide a figure into 2 right triangles
\nWe know that,
\nWe can apply the Pythagorean Theorem for any right triangle
\nWe know that,
\nIn a right triangle,
\nThe hypotenuse is the longest side
\nWe know that,
\nThe hypotenuse in a right triangle is considered as a diagonal in a figure that consists of 1 right angle
\nHence, from the above,
\nWe can conclude that Tim is correct<\/p>\n

Essential Question<\/strong>
\nWhat types of problems can be solved using the Pythagorean Theorem?
\nAnswer:
\nThe Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the sides as long as you know the lengths of the other two sides<\/p>\n

Try It!<\/strong><\/p>\n

What is the length of the diagonal, d, of a rectangle with length 19 feet and width 17 feet?
\nleg2<\/sup> + leg2<\/sup> = hypotenuse2<\/sup>
\n______ 2<\/sup> + _______2<\/sup> = d2<\/sup>
\n______ + _______ = d2<\/sup>
\n_______ = d2<\/sup>
\n________ \u2248 d
\nAnswer:
\nIt is given that a rectangle has a length of 19 feet and a width of 17 feet
\nNow,
\nWe know that,
\nIn a rectangle,
\nIf a diagonal is drawn, then it divides the rectangle into 2 right angles
\nSo,
\nAccording to the Pythagorean Theorem,
\nd\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nd is the diagonal or hypotenuse
\na and b are the lengths of the legs
\nSo,
\nd\u00b2= 19\u00b2 + 17\u00b2
\nd\u00b2 = 361 + 289
\nd\u00b2 = 650
\nd = \\(\\sqrt{650}\\)
\nd = 25.4 ft
\nHence, from the above,
\nWe can conclude that the length of the diagonal is: 25.4 ft<\/p>\n

Convince Me!
\nIf the rectangle were a square, would the process of finding the length of the diagonal change? Explain.
\nAnswer:
\nWe know that,
\nFor any figure i.e., either 2-d figure or 3-d figure with 1 right angle, the diagonal will divide that figure into 2 right angles
\nWe know that,
\nWe will use the Pythagorean Theorem to find the length of any unknown side in the right triangle
\nHence, from the above,
\nWe can conclude that even if the rectangle were a square, the process of finding the length of the diagonal will not change<\/p>\n

Try It!<\/strong><\/p>\n

A company wants to rent a tent that has a height of at least 10 feet for an outdoor show. Should they rent the tent shown at the right? Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\nA company wants to rent a tent that has a height of at least 10 feet for an outdoor show.
\nNow,
\nThe given figure is:
\n\"Envision
\nNow,
\nTo find whether the tent should be rented or not,
\nWe have to find the value of h
\nWe can also observe that the triangle that contains the value of h is a right triangle
\nNow,
\nThe base of the right triangle = \\(\\frac{24}{2}\\)
\n= 12 ft
\nNow,
\nThe side lengths of the right triangle are: h, 15 ft, 12 ft
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nSo,
\n15\u00b2 = 12\u00b2 + h\u00b2
\nh\u00b2 = 225 – 144
\nh\u00b2 = 81
\nh = \\(\\sqrt{81}\\)
\nh = 9 ft
\nBut,
\nThe given height of the tent is: 10 ft
\nSo,
\n9 ft < 10 ft
\nHence, from the above,
\nWe can conclude that the company should not rent the tent<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

You can use the Pythagorean Theorem and its converse to solve problems involving right triangles.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> What types of problems can be solved using the Pythagorean Theorem?
\nAnswer:
\nThe Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the sides as long as you know the lengths of the other two sides<\/p>\n

Question 2.
\nLook for Structure How is using the Pythagorean Theorem in a rectangular prism similar to using it in a rectangle?
\nAnswer:
\nWe know that,
\nThe rectangular prism and the rectangle have at least 1 right angle
\nWe know that,
\nIf a 3-d or 2-d figure has 1 right angle, then the diagonal of that figure divides the figure into the right triangles
\nSo,
\nIf we have the right triangle, then we can use the Pythagorean Theorem irrespective of the overall shape of the figure<\/p>\n

Question 3.
\nConstruct Arguments<\/strong> Glen found the length of the hypotenuse of a right triangle using \\(\\sqrt{a^{2}+b^{2}}\\). Gigi used \\(\\sqrt{(a+b)^{2}}\\). Who is correct? Explain.
\nAnswer:
\nIt is given that
\nGlen found the length of the hypotenuse of a right triangle using \\(\\sqrt{a^{2}+b^{2}}\\). Gigi used \\(\\sqrt{(a+b)^{2}}\\).
\nNow,
\nWe know that,
\nWe can use the Pythagorean Theorem only for the right triangles
\nThe condition for a triangle to become the right triangle is:
\nHypotenuse\u00b2 = Side\u00b2 + Side\u00b2
\nc\u00b2 = a\u00b2 + b\u00b2
\nHence, from the above,
\nWe can conclude that Glen is correct<\/p>\n

Do You Know How?<\/strong>
\nQuestion 4.
\nYou are painting the roof of a boathouse. You are going to place the base of a ladder 12 feet from the boathouse. How long does the ladder need to be to reach the roof of the boathouse?
\n\"Envision
\nAnswer:
\nIt is given that
\nYou are painting the roof of a boathouse. You are going to place the base of a ladder 12 feet from the boathouse.
\nNow,
\nFrom the figure,
\nWe can observe that the roof of a boathouse, ladder, and the base form a right triangle
\nSo,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the height of the boathouse
\na is the length of the base
\nb is the length of the ladder
\nSo,
\n35\u00b2 = 12\u00b2 + b\u00b2
\nb\u00b2 = 1,225 – 144
\nb\u00b2 = 1,081
\nb = \\(\\sqrt{1,081}\\)
\nb = 32.9 ft
\nHence, from the above,
\nWe can conclude that the length of the ladder that is needed to reach the roof of the boathouse is: 32.9 ft<\/p>\n

Question 5.
\nA box-shaped like a right rectangular prism measures 5 centimeters by 3 centimeters by 2 centimeters. What is the length of the interior diagonal of the prism to the nearest hundredth?
\nAnswer:
\nIt is given that
\nA box-shaped like a right rectangular prism measures 5 centimeters by 3 centimeters by 2 centimeters.
\nSo,
\nThe dimensions of a rectangular prism is: 5 cm \u00d7 3 cm \u00d7 2 cm
\nSo,
\nThe length of the rectangular prism is: 5 cm
\nThe width of the rectangular prism is: 3 cm
\nThe height of the rectangular prism is: 2 cm
\nNow,
\nWe know that,
\nThe length of the diagonal of the prism = \\(\\sqrt{Length^{2} + Width^{2} + Height^{2}}\\)
\nSo,
\nThe length of the diagonal of the rectangular prism = \\(\\sqrt{5^{2} + 3^{2} + 2^{2}}\\)
\n= \\(\\sqrt{25 + 9 + 4}\\)
\n= \\(\\sqrt{38}\\)
\n= 6.16 cm
\nHence, from the above,
\nWe can conclude that the length of the interior diagonal of the rectangular prism is: 6.16 cm<\/p>\n

Question 6.
\nA wall 12 feet long makes a corner with a wall that is 14 feet long. The other ends of the walls are about 18.44 feet apart. Do the walls form a right angle? Explain.
\n\"\"
\nAnswer:
\nIt is given that
\nA wall 12 feet long makes a corner with a wall that is 14 feet long. The other ends of the walls are about 18.44 feet apart
\nNow,
\nThe given figure is:
\n\"\"
\nFrom the given figure,
\nWe can observe that the given situation forms the right triangle
\nNow,
\nWe know that,
\nAccording to the Pythagorean Therem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nSo,
\nc\u00b2 = 12\u00b2 + 14\u00b2
\nc\u00b2 = 144 + 196
\nc\u00b2 = 340
\nc = \\(\\sqrt{340}\\)
\nc = 18.44 feet
\nHence, from the above,
\nWe can conclude that the walls form a right angle<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Leveled Practice In 7 and 8, use the Pythagorean Theorem to solve.
\nQuestion 7.
\nYou are going to use an inclined plane to lift a heavy object to the top of a shelving unit with a height of 6 feet. The base of the inclined plane is 16 feet from the shelving unit. What is the length of the inclined plane? Round to the nearest tenth of a foot.
\n\"Envision
\nThe length of the inclined plane is about ________ feet.
\nAnswer:
\n\"\"<\/p>\n

Question 8.
\nFind the missing lengths in the rectangular prism.
\n\"Envision
\nAnswer:
\n\"\"<\/p>\n

Question 9.
\nA stainless steel patio heater is shaped like a square pyramid. The length of one side of the base is 19.8 inches. The slant height is 92.8 inches. What is the height of the heater? Round to the nearest tenth of an inch.
\nAnswer:
\nIt is given that
\nA stainless steel patio heater is shaped like a square pyramid. The length of one side of the base is 19.8 inches. The slant height is 92.8 inches
\nNow,
\nWe know that,
\nThe slant height is nothing but a length of the diagonal
\nNow,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the slant height
\na is the length of the base of the steel patio heater
\nb is the height of the heater
\nSo,
\n(92.8)\u00b2 = (19.8)\u00b2 + b\u00b2
\nb\u00b2 = 8,611.84 – 392.04
\nc\u00b2 = 8,219.8
\nc = \\(\\sqrt{8,219.8}\\)
\nc = 90.6 inches
\nHence, from the above,
\nWe can conclude that the height of the heater is: 90.6 inches<\/p>\n

Question 10.
\nReasoning<\/strong> What is the measurement of the longest line segment in a right rectangular prism that is 16 centimeters long, 9 centimeters wide, and 7 centimeters tall? Round to the nearest tenth of a centimeter.
\nAnswer:
\nIt is given that
\nA right rectangular prism is 16 centimeters long, 9 centimeters wide, and 7 centimeters tall
\nSo,
\nThe dimensions of a right rectangular prism are: 16 cm \u00d7 9 cm \u00d7 7 cm
\nSo,
\nThe length of the right rectangular prism is: 16 cm
\nThe width of the right rectangular prism is: 9 cm
\nThe height of the right rectangular prism is: 7 cm
\nWe know that,
\nThe longest line segment in any 2-d or 3-d figure is a “Diagonal”
\nNow,
\nWe know that,
\nThe length of the diagonal of the right rectangular prism = \\(\\sqrt{Length^{2} + Width^{2} + Height^{2}}\\)
\nSo,
\nThe length of the diagonal of the right rectangular prism = \\(\\sqrt{16^{2} + 9^{2} + 7^{2}}\\)
\n= \\(\\sqrt{256 + 49 + 81}\\)
\n= \\(\\sqrt{386}\\)
\n= 19.64 cm
\nHence, from the above,
\nWe can conclude that the length of the longest line segment in the right rectangular prism is: 19.64 cm<\/p>\n

Question 11.
\nFelipe is making triangles for a stained glass window. He made the design shown but wants to change it. Felipe wants to move the purple triangle to the corner. The purple piece has side lengths of 4.5 inches, 6 inches, and 7 inches. Can the purple piece be moved to the corner? Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\nFelipe is making triangles for a stained glass window. He made the design shown as above but wants to change it. Felipe wants to move the purple triangle to the corner. The purple piece has side lengths of 4.5 inches, 6 inches, and 7 inches.
\nNow,
\nFrom the given figure,
\nWe can observe that
\nIf the purple figure is moving to the corner, then one side of the purple figure will become a right angle
\nSo,
\nNow,
\nWe have to find whether the purple figure will be a right triangle or not if it moves to a corner
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2,
\nthen, the triangle is a right triangle
\nWe know that,
\nThe hypotenuse has the longest side length
\nSo,
\n7\u00b2 = 4.5\u00b2 + 6\u00b2
\n49 = 20.25 + 36
\n49 = 56.25
\nSo,
\nThe condition
\nc\u00b2 = a\u00b2 + b\u00b2
\nis false
\nHence, from the above,
\nWe can conclude that the purple figure should not<\/p>\n

Question 12.
\na. What is the longest poster you could fit in the box? Express your answer to the nearest tenth of an inch.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nWe can observe that there will be two longest sides for two pairs of different side lengths
\nwe know that,
\nThe longest side is nothing but a diagonal
\nNow,
\nThe two pairs of side lengths are: (20, 12, x), and (8, 12, y)
\nWhere,
\nx and y are the lengths of the longest sides
\nNow,
\nWe know that
\nSince the figure consists of the right triangles,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nSo,
\nx\u00b2 = 20\u00b2 + 12\u00b2
\nx\u00b2 = 400 + 144
\nx\u00b2 = 544
\nx = \\(\\sqrt{544}\\)
\nx = 23.32 in.
\nSo,
\ny\u00b2 = 12\u00b2 + 8\u00b2
\ny\u00b2 = 144 + 64
\ny\u00b2 = 208
\ny = \\(\\sqrt{208}\\)
\ny = 14.42 in.
\nHence, from the above,
\nWe can conclude that the longest poster that you can fit in the box is: 23.32 in.<\/p>\n

b. Explain why you can fit only one maximum-length poster in the box, but you can fit multiple 21.5-inch posters in the same box.
\n\"Envision
\nAnswer:
\nIt is given that
\nyou can fit multiple 21.5-inch posters in the same box.
\nNow,
\nFrom part (a),
\nWe can observe that the length of the longest poster that can fit into the box is: 23.32 in.
\nSo,
\n21.5 in. < 23.32 in.
\nHence, from the above,
\nWe can conclude that since the given size of the posters is less than the maximum length of the poster,
\nWe can fit multiple 21.5-inch posters in the same box instead of 1 poster that is of the maximum length<\/p>\n

Question 13.<\/span><\/p>\n

The corner of a room where two walls meet the floor should be at a right angle. Jeff makes a mark along each wall. One mark is 3 inches from the corner. The other is 4 inches from the corner. How can Jeff use the Pythagorean Theorem to see if the walls form a right angle?
\nAnswer:
\nIt is given that
\nThe corner of a room where two walls meet the floor should be at a right angle. Jeff makes a mark along each wall. One mark is 3 inches from the corner. The other is 4 inches from the corner.
\nNow,
\nTo see whether the walls form a right angle or not,
\nWe have to see whether the length along the walls is greater than the lengths of the marks from the corners
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length along the walls
\na is the length of one mark from the corner
\nb is the length of another mark from the corner
\nSo,
\nc\u00b2 = 3\u00b2 + 4\u00b2
\nc\u00b2 = 9 + 16
\nc\u00b2 = 25
\nc = \\(\\sqrt{25}\\)
\nc = 5
\nSo,
\nFrom the above value,
\nWe can observe that
\nc > a and c > b
\nHence, from the above,
\nWe can conclude that the walls form a right angle<\/p>\n

Question 14.
\nHigher-Order Thinking<\/strong> It is recommended that a ramp have at least 6 feet of horizontal distance for every 1 foot of vertical rise along an incline. The ramp shown has a vertical rise of 2 feet. Does the ramp show match the recommended specifications? Explain.
\n\"Envision
\nAnswer:
\nIt is given that
\nIt is recommended that a ramp have at least 6 feet of horizontal distance for every 1 foot of vertical rise along an incline. The ramp shown has a vertical rise of 2 feet
\nNow,
\n6 feet of horizontal distance for every 1 foot,
\nThe ramp shown has a vertical rise of 5 feet.
\nSo,
\nThe rate of change (m)\u00a0 \u2264 \\(\\frac{1 foot}{6 feet}\\)
\nConvert both to a single unit of inches
\nWe know that,
\n1 foot = 12 inches
\nSo,
\nm \u2264 \\(\\frac{12 inches}{72 inches}\\)
\nDivide by 4 into both sides
\nm \u2264 \\(\\frac{4 inches}{18 inches}\\)
\nNow,
\nFor a ramp that has a vertical distance of 2 feet and the same horizontal distance
\nm = \\(\\frac{12 inches}{24 inches}\\)
\nm = \\(\\frac{4 inches}{6 inches}\\)
\n<\/span>So,
\nThe rate of change when the vertical distance is 6 feet > The rate of change when the vertical distance is 2 feet
\nHence, from the above,
\nWe can conclude that the ramp shown above matched the given specifications<\/p>\n

Assessment Practice<\/strong><\/p>\n

Question 15.
\nA machine in a factory cuts out triangular sheets of metal. Which of the triangles are right triangles? Select all that apply.
\n\"Envision
\n\u2610 Triangle 1
\n\u2610 Triangle 2
\n\u2610 Triangle 3
\n\u2610 Triangle 4
\nAnswer:
\nIt is given that
\nA machine in a factory cuts out triangular sheets of metal.
\nNow,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2
\nthen, the triangle is a right triangle
\nNow,
\nFor Triangle 1,
\n(\\(\\sqrt{505}\\))\u00b2 = 12\u00b2 + 19\u00b2
\n505 = 505
\nFor Triangle 2,
\n(\\(\\sqrt{467}\\))\u00b2 = 16\u00b2 + 19\u00b2
\n467 \u2260 617
\nFor Triangle 3,
\n(\\(\\sqrt{596}\\))\u00b2 = 14\u00b2 + 20\u00b2
\n596 = 596
\nFor Triangle 4,
\n(\\(\\sqrt{421}\\))\u00b2 = 11\u00b2 + 23\u00b2
\n421 \u2260 650
\nHence, from the above,
\nWe can conclude that Triangle 1 and Triangle 3 are the right triangles<\/p>\n

Question 16.
\nWhat is the length b, in feet, of the rectangular plot of land shown?
\n\"Envision
\nAnswer:
\nIt is given that the given figure is a rectangular plot of land
\nWe know that,
\nIn a rectangle,
\nThe opposite sides are equal and the angles are all 90\u00b0
\nSo,
\nA diagonal forms 2 right triangles in a rectangle
\nNow,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the length of the diagonal
\na is the width
\nb is the length
\nNow,
\n325\u00b2 = 300\u00b2 + b\u00b2
\nb\u00b2= 1,05,625 – 90,000
\nb\u00b2 = 15,625
\nb = \\(\\sqrt{15,625}\\)
\nb = 125 ft
\nHence, from the above,
\nWe can conclude that the length of b is: 125 ft<\/p>\n

Lesson 7.4 Find Distance in the Coordinate<\/h3>\n

Explore It!<\/strong>
\nThomas and Jim are outside the haunted castle ride and want to get to the clown tent in time for the next show.
\n\"Envision
\nI can… use the Pythagorean Theorem to find the distance between two points in the coordinate plane.<\/p>\n

A. How can you represent the starred locations on a coordinate plane?
\n\"Envision
\nAnswer:
\nIt is given that
\nThomas and Jim are outside the haunted castle ride and want to get to the clown tent in time for the next show.
\nNow,
\nThe given route of the castle is:
\n\"Envision
\nNow,
\nFrom the given figure,
\nTo represent the route in the coordinate plane,
\nThe given scale is:
\nX-axis: 1 cm = 500 feet
\nY-axis: 1 cm = 500 feet
\nNow,
\nFrom the given figure,
\nThe coordinates of the starred locations can be:
\nThe coordinates of the haunted house are: (500, 1,500),
\nThe coordinates of the clown tent are: (2,000, 500)
\nHence,
\nThe representation of the starred locations in the coordinate plane is:
\n\"\"<\/p>\n

B. Jim says that the marked yellow paths show the shortest path to the tent. Write an expression to represent this and find the distance Jim walks from the haunted mansion to the clown tent.
\nAnswer:
\nFrom part (a),
\nWe know that,
\nThe coordinates of the haunted house are: (500, 1,500),
\nThe coordinates of the clown tent are: (2,000, 500)
\nNow,
\nWe know that
\nThe linear equation in the slope-intercept form is:
\ny = mx + b
\nWhere,
\nm is the slope
\nb is the initial value (or) y-intercept
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nSlope (m) = y2<\/sub> – y1<\/sub> \/ x2<\/sub> – x1<\/sub>
\nSo,
\nSlope (m) = \\(\\frac{500 – 1,500}{2,000 – 500}\\)
\nSlope (m) = –\\(\\frac{1,000}{1,500}\\)
\nSlope (m) = –\\(\\frac{2}{3}\\)
\nSo,
\nThe equation in the slope-intercept form is:
\ny = –\\(\\frac{2}{3}\\)x + b
\nSo,
\n3y = -2x + 3b
\nNow,
\nSubstitute (500, 1,500) or (2,000, 500) in the above equation
\nSo,
\n1,500 = -4000 + 3b
\n3b = 5,500
\nSo,
\nThe equation that represents the shortest path to the tent is:
\n3y = -2x + 5,500
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the haunted mansion and the clown tent = \\(\\sqrt{(2,000 – 500)^{2} + (500 – 1,500)^{2}}\\)
\n= \\(\\sqrt{1,500^{2} + 1,000^{2}}\\)
\n= \\(\\sqrt{2,250,000 + 1,000,000}\\)
\n= 1,802.77 feet
\nHence, from the above,
\nWe can conclude that
\nThe equation that represents the shortest path to the clown tent is:
\n3y = -2x + 5,500
\nThe distance between the haunted mansion and the clown tent is: 1,802.77 feet<\/p>\n

Focus on math practices
\nConstruct Arguments Why is the distance between two nonhorizontal and nonvertical points always greater than the horizontal or vertical distance?
\nAnswer:
\nLet us consider a coordinate plane
\nNow,
\nWhen we draw either a horizontal line or the vertical line,
\nWe can observe that the length will be constant
\nBut,
\nWhen we draw non-vertical and non-horizontal lines,
\nWe can observe that the lengths are unknown and not constant
\nHence, from the above,
\nWe can conclude that the distance between two nonhorizontal and nonvertical points always greater than the horizontal or vertical distance<\/p>\n

Essential Question<\/strong>
\nHow can you use the Pythagorean Theorem to find the distance between two points?
\nAnswer:
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the side lengths or lengths of the legs
\nNow,
\nGraphically,
\nThe terms of the Pythagorean Theorem can be expressed as:
\nc is the distance between two points
\na and b are the points
\nSo,
\nc = \\(\\sqrt{a^{2} + b^{2}}\\)<\/p>\n

Try It!<\/strong><\/p>\n

What is the distance between points A and B?
\nThe distance between point A and point B is about ________ units.
\n\"Envision
\nAnswer:
\nThe given graph is:
\n\"Envision
\nThe representation of the graph in the coordinate plane is:
\n\"\"
\nFrom the given graph,
\nThe coordinates of A are: (2, 3)
\nThe coordinates of B are: (4, 1)
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points A and B = \\(\\sqrt{(4 – 2)^{2} + (1 – 3)^{2}}\\)
\n= \\(\\sqrt{2^{2} + 2^{2}}\\)
\n= \\(\\sqrt{4 + 4}\\)
\n= 2.83 units
\nHence, from the above,
\nWe can conclude that the distance between points A and B is about 2.83 units<\/p>\n

Convince Me!
\nWhy do you need to use the Pythagorean Theorem to find the distance between points A and B?
\nAnswer:
\nThe representation of points A and B in the coordinate plane are:
\n\"\"
\nNow,
\nWhen we observe the graph,
\nWe can see that A and B can form a right triangle
\nNow,
\nWe know that,
\nThe Pythagorean Theorem is only applicable to the right triangles
\nSo,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nc = \\(\\sqrt{a^{2} + b^{2}}\\)
\nwhere,
\nc is the distance between points A and B
\nA and B are the given points<\/p>\n

Try It!
\n<\/strong>Find the perimeter of \u2206ABC with vertices (2, 5), (5, -1), and (2, -1).
\nAnswer:
\nIt is given that
\n\u2206ABC with vertices (2, 5), (5, -1), and (2, -1)
\nNow,
\nThe names of the vertices are:
\nA (2, 5), B (5, -1), and C (2, -1)
\nWe know that,
\nThe perimeter of a triangle is the sum of all the side lengths of a triangle
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nIn \u2206ABC,
\nAB and BC are the side lengths
\nAc is the hypotenuse
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points A and B = \\(\\sqrt{(5 – 2)^{2} + (-1 – 5)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 6^{2}}\\)
\n= \\(\\sqrt{9 + 36}\\)
\n= 6.70 units
\nThe distance between the points B and C (BC) = \\(\\sqrt{(2 – 5)^{2} + (-1 + 1)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 0^{2}}\\)
\n= \\(\\sqrt{9 + 0}\\)
\n= 3 units
\nThe distance between the points A and C (AC) = \\(\\sqrt{(2 – 2)^{2} + (-1 – 5)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 6^{2}}\\)
\n= \\(\\sqrt{0 + 36}\\)
\n= 6 units
\nSo,
\nThe perimeter of \u2206ABC = AB + BC + AC
\n= 6 + 3 + 6.70
\n= 15.7 units
\nHence, from the above,
\nWe can conclude that the perimeter of \u2206ABC is about 15.7 units<\/p>\n

Try It!<\/strong><\/p>\n

What are the coordinates, to the nearest tenth, of the third vertex in an isosceles triangle that has one side length of 2 and two side lengths of 5, with vertices at (1, 0) and (1, 2)? The third vertex is in the first quadrant.
\nAnswer:
\nIt is given that
\nAn isosceles triangle that has one side length of 2 and two side lengths of 5, with vertices at (1, 0) and (1, 2)
\nNow,
\nLet the third vertex be (x, y)
\nNow,
\nThe given vertices are:
\nA (1, 0), B (1, 2), and C (x, y)
\nIt is also given that
\nBC = 2 units, and AC = 5 units
\nWe know that,
\nAn isosceles triangle has any 2 equal side lengths
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points A and B = \\(\\sqrt{(2 – 0)^{2} + (1 – 1)^{2}}\\)
\n= \\(\\sqrt{2^{2} + 0^{2}}\\)
\n= \\(\\sqrt{4 + 0}\\)
\n= 4 units
\nThe distance between the points B and C = \\(\\sqrt{(x – 1)^{2} + (y – 2)^{2}}\\)
\nSquaring on both sides
\nSo,
\nBC\u00b2 = (x – 1)\u00b2 + (y – 2)\u00b2
\nThe distance between the points A and C = \\(\\sqrt{(x – 1)^{2} + (y – 0)^{2}}\\)
\nSquaring on both sides
\nSo,
\nBC\u00b2 = (x – 1)\u00b2 + y\u00b2
\nSo,
\n(x – 1)\u00b2 + (y – 2)\u00b2 = 4 —– (1)
\n(x – 1)\u00b2 + y\u00b2 = 25 —— (2)
\nSo,
\nFrom eq (1) and eq (2),
\n25 – y\u00b2 + (y – 2)\u00b2 = 4
\n-y\u00b2 + y\u00b2 – 4y + 4 = -21
\n-4y = -25
\ny = \\(\\frac{25}{4}\\)
\nSo,
\n(x – 1)\u00b2 = |25 – (\\(\\frac{25}{4}\\))\u00b2|
\nx = \\(\\frac{19}{4}\\)
\nHence, from the above,
\nWe can conclude that the third vertex is: (\\(\\frac{19}{4}\\), \\(\\frac{25}{4}\\))<\/p>\n

KEY CONCEPT<\/strong><\/p>\n

You can use the Pythagorean Theorem to find the distance between any two points, P and Q, on the coordinate plane.
\n\"Envision<\/p>\n

Do You Understand?<\/strong>
\nQuestion 1.
\nEssential Question<\/strong> How can you use the Pythagorean Theorem to find the distance between two points?
\nAnswer:
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the side lengths or lengths of the legs
\nNow,
\nGraphically,
\nThe terms of the Pythagorean Theorem can be expressed as:
\nc is the distance between two points
\na and b are the points
\nSo,
\nc = \\(\\sqrt{a^{2} + b^{2}}\\)<\/p>\n

Question 2.
\nModel with Math<\/strong>
\nCan you use a right triangle to represent the distance between any two points on the coordinate plane? Explain.
\nAnswer:
\nDerived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. We know that,
\nAccording to the Pythagorean Theorem,
\na\u00b2+b\u00b2=c\u00b2
\nwhere,
\na and b are the lengths of the legs adjacent to the right angle
\nc is the length\u00a0of\u00a0the hypotenuse.<\/p>\n

Question 3.
\nGeneralize<\/strong> How does the fact that the points are on opposite sides of the y-axis affect the process of finding the distance between the two points?
\nAnswer:
\nThe fact that the points are on opposite sides of the y-axis affects the process of finding the distance between the two points because\u00a0 We need\u00a0to find the distance between the two points\u00a0by adding the\u00a0distances from\u00a0each of them\u00a0to\u00a0the\u00a0y-axis.<\/p>\n

Do You Know How?<\/strong>
\nIn 4-6, use the coordinate plane below.
\n\"Envision
\nAnswer:
\nThe given coordinate plane is:
\n\"Envision
\nFrom the given coordinate plane,
\nThe given points are:
\nC (1, 2), D (2, -1), and E (-2, 1)<\/p>\n

Question 4.
\nFind the distance between points C and D. Round to the nearest hundredth.
\nAnswer:
\nCompare the points C and D with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points C and D = \\(\\sqrt{(2 – 1)^{2} + (-1 – 2)^{2}}\\)
\n= \\(\\sqrt{1^{2} + 3^{2}}\\)
\n= \\(\\sqrt{1 + 9}\\)
\n= 3.16 units
\nHence, from the above,
\nWe can conclude that the distance between points C and D is: 3.16 units<\/p>\n

Question 5.
\nFind the perimeter of \u2206CDE.
\nAnswer:
\nWe know that,
\nThe “Perimeter” is defined as the sum of all the side lengths
\nSo,
\nThe perimeter of \u2206CDE = CD + DE + CE
\nSo,
\nThe distance between the points C and D = \\(\\sqrt{(2 – 1)^{2} + (-1 – 2)^{2}}\\)
\n= \\(\\sqrt{1^{2} + 3^{2}}\\)
\n= \\(\\sqrt{1 + 9}\\)
\n= 3.16 units
\nThe distance between the points D and E = \\(\\sqrt{(-2 – 2)^{2} + (1 + 1)^{2}}\\)
\n= \\(\\sqrt{4^{2} + 2^{2}}\\)
\n= \\(\\sqrt{16 + 4}\\)
\n= 4.47 units
\nThe distance between the points C and E = \\(\\sqrt{(-2 – 1)^{2} + (1 – 2)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 1^{2}}\\)
\n= \\(\\sqrt{1 + 9}\\)
\n= 3.16 units
\nSo,
\nThe perimeter of \u2206CDE = 3.16 + 4.47 + 3.16
\n= 10.79 units
\nHence, from the above,
\nWe can conclude that the perimeter of \u2206CDE is: 10.79 units<\/p>\n

Question 6.
\nPoint B is plotted on the coordinate plane above the x-axis. \u2206BDE is equilateral. What are the coordinates of point B to the nearest hundredth?
\nAnswer:
\nIt is given that
\nPoint B is plotted on the coordinate plane above the x-axis. \u2206BDE is equilateral.
\nNow,
\nLet the unknown vertex be B (x, y)
\nSo,
\nThe given points are:
\nB (x, y), D (2, -1), and E (-2, 1)
\nIt is given that \u0394BDE is equilateral
\nSo,
\nBD = DE = EB
\nBD\u00b2 = DE\u00b2 = EB\u00b2
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points B and D = \\(\\sqrt{(x – 2)^{2} + (y – 1)^{2}}\\)
\nSquaring on both sides
\nSo,
\nBD\u00b2 = (x – 2)\u00b2 + (y + 1)\u00b2
\nThe distance between the points D and E = \\(\\sqrt{(1 + 1)^{2} + (-2 – 2)^{2}}\\)
\n= \\(\\sqrt{2^{2} + 4^{2}}\\)
\n= \\(\\sqrt{4 + 16}\\)
\n= 4.47 units
\nThe distance between the points E and B = \\(\\sqrt{(x + 2)^{2} + (y – 1)^{2}}\\)
\nSquaring on both sides
\nSo,
\nEB\u00b2 = (x + 2)\u00b2 + (y – 1)\u00b2
\nNow,
\n(x – 2)\u00b2 + (y + 1)\u00b2 = 4.47 —- (1)
\n(x + 2)\u00b2 + (y – 1)\u00b2 = (x – 2)\u00b2 + (y + 1)\u00b2
\nx\u00b2 + 2x + 4 + y\u00b2 + 1 – 2y = x\u00b2 + 4 – 4x + y\u00b2 + 1 + 2y
\n2x + 4 + 1 – 2y = 4 – 4x + 1 + 2y
\n6x + 5 = 4y + 5
\n6x = 4y
\n3x = 2y
\nx = \\(\\frac{2}{3}\\)y
\nNow,
\nFrom eq (1),
\nx\u00b2 + 4 – 4x + y\u00b2 + 1 + 2y = 4.47
\nx\u00b2 + y\u00b2 -4x + 2y = -0.47
\n(\\(\\frac{2}{3}\\)y)\u00b2 + y\u00b2 – 4 (\\(\\frac{2}{3}\\))y + 2y = -0.47
\n4y\u00b2 + 15y + 4.23 = 0
\nSo,
\ny = -0.30 (or) y = -3.44
\nSo,
\nx = \\(\\frac{2}{3}\\) (-0.30) (or) x = \\(\\frac{2}{3}\\) (-3.44)
\nx = -0.2 (or) x = -2.29
\nHence, from the above,
\nWe can conclude that the coordinates of point B are: (-0.2, -0.30) or (-2.29, -3.44)<\/p>\n

Practice & Problem Solving<\/strong><\/p>\n

Question 7.
\nLeveled Practice<\/strong> Use the Pythagorean Theorem to find the distance between points P and Q.
\nLabel the length, in units, of each leg of the right triangle.
\n\"Envision
\nThe distance between point P and point Q is __________ units.
\nAnswer:
\nFrom the given coordinate plane,
\nThere are only 2 vertices
\nLet the third vertex be R(x, y) and the coordinates of R can be found from the coordinate plane
\nNow,
\nFrom the given coordinate plane,
\nThe vertices are:
\nP (3, 2), Q (9, 10), and R (9, 2)
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points Q and R = \\(\\sqrt{(9 – 9)^{2} + (10 – 2)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 8^{2}}\\)
\n= \\(\\sqrt{0 + 64}\\)
\n= 8 units
\nThe distance between the points P and R = \\(\\sqrt{(9 – 3)^{2} + (2 – 2)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 6^{2}}\\)
\n= \\(\\sqrt{0 + 36}\\)
\n= 6 units
\nNow,
\nFrom the given coordinate plane,
\nWe can observe that P, Q, R form the right triangle
\nSo,
\nAccording to the Pythagorean Theorem,
\nPQ\u00b2 = QR\u00b2 + PR\u00b2
\nPQ\u00b2 = 8\u00b2 + 6\u00b2
\nPQ\u00b2 = 64 + 36
\nPQ\u00b2 = 100
\nPQ = \\(\\sqrt{100}\\)
\nPQ = 10 units
\nHence, from the above,
\nWe can conclude that the length of PQ is: 10 units<\/p>\n

\"Envision
\nQuestion 8.
\nFind the perimeter of triangle QPR. Round to the nearest hundredth.
\nAnswer:
\nFrom the coordinate plane,
\nThe vertices of \u0394PQR are:
\nP (-5, -2), Q (2, -2), and R (-1, 3)
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points P and Q = \\(\\sqrt{(2 + 5)^{2} + (2 – 2)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 7^{2}}\\)
\n= \\(\\sqrt{0 + 49}\\)
\n= 7 units
\nThe distance between the points Q and R = \\(\\sqrt{(-1 – 2)^{2} + (3 + 2)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 5^{2}}\\)
\n= \\(\\sqrt{9 + 25}\\)
\n= 5.83 units
\nThe distance between the points P and R = \\(\\sqrt{(1 – 5)^{2} + (-3 – 2)^{2}}\\)
\n= \\(\\sqrt{4^{2} + 5^{2}}\\)
\n= \\(\\sqrt{16 + 25}\\)
\n= 6.40 units
\nSo,
\nThe perimeter of \u0394PQR = PQ + QR + PR
\n= 7 + 5.83 + 6.40
\n= 19.23 units
\nHence, from the above,
\nWe can conclude that the perimeter of \u0394PQR is: 19.23 units<\/p>\n

Question 9.
\nDetermine whether the triangle is equilateral, isosceles, or scalene.
\nAnswer:
\nWe know that,
\nOn the basis of the side lengths,
\nScalene Triangle – All the side lengths are different
\nEquilateral Triangle – All the side lengths are the same
\nIsosceles Triangle – Any two of the side lengths are the same
\nSo,
\nFrom Exercise 8,
\nWe can observe that all the side lengths are different
\nHence, from the above,
\nWe can conclude that \u0394PQR is a scalene Triangle<\/p>\n

Question 10.
\nYou walk along the outside of a park starting at point P. Then you take a shortcut represented by \\(\\overline{P Q}\\) on the graph.
\n\"Envision
\na. What is the length of the shortcut in meters? Round to the nearest tenth of a meter.
\nAnswer:
\nIt is given that
\nYou walk along the outside of a park starting at point P. Then you take a shortcut represented by \\(\\overline{P Q}\\) on the graph.
\nNow,
\nFrom the given figure,
\nWe can observe that
\nThe vertices are:
\nP (0, 0), Q (40, 85), and R (40, 0)
\nThe shortest path is represented as PQ
\nNow,
\nCompare the points P and Q with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points P and Q = \\(\\sqrt{(40 – 0)^{2} + (85 – 0)^{2}}\\)
\n= \\(\\sqrt{40^{2} + 85^{2}}\\)
\n= \\(\\sqrt{1,600 + 7,225}\\)
\n= 93.9 m
\nHence, from the above,
\nWe can conclude that the length of the shortest path is: 93.9 meters<\/p>\n

b. What is the total length of your walk in the park? Round to the nearest tenth of a meter.
\nAnswer:
\nWe know that,
\nThe total length is nothing but the “Perimeter”
\nSo,
\nThe perimeter of the given triangle = PQ + QR + PR
\nNow,
\nThe distance between the points P and Q = \\(\\sqrt{(40 – 0)^{2} + (85 – 0)^{2}}\\)
\n= \\(\\sqrt{40^{2} + 85^{2}}\\)
\n= \\(\\sqrt{1,600 + 7,225}\\)
\n= 93.9 m
\nThe distance between the points Q and R = \\(\\sqrt{(40 – 40)^{2} + (85 – 0)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 85^{2}}\\)
\n= \\(\\sqrt{0 + 7,225}\\)
\n= 85 m
\nThe distance between the points P and R = \\(\\sqrt{(40 – 0)^{2} + (0 – 0)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 40^{2}}\\)
\n= \\(\\sqrt{0 + 1,600}\\)
\n= 40 m
\nSo,
\nThe total length of your walk in the park = PQ + QR + PR
\n= 93.9 + 85 + 40
\n= 218.9 meters
\nHence, from the above,
\nWe can conclude that the total length of your walk in the park is: 218.9 meters<\/p>\n

Question 11.
\nSuppose a park is located 3.6 miles east of your home. The library is 4.8 miles north of the park. What is the shortest distance between your home and the library?
\nAnswer:
\nIt is given that
\nA park is located 3.6 miles east of your home. The library is 4.8 miles north of the park
\nSo,
\nThe representation of the above situation is:
\n\"\"
\nNow,
\nAccording to the Pythagorean Theorem,
\n(The shortest distance between home and library)\u00b2 = (The distance from home to park)\u00b2 + (he distance from park to library)\u00b2
\n(The shortest distance between home and library)\u00b2 = 3.6\u00b2 + 4.8\u00b2
\n(The shortest distance between home and library)\u00b2 = 12.96 + 23.04
\nThe shortest distance between home and library = 6
\nHence, from the above,
\nWe can conclude that the shortest distance between home and library is: 6 miles<\/p>\n

Question 12.
\nUse Structure Point B has coordinates (2, 1). The x-coordinate of point A is -10. The distance between point A and point B is 15 units. What are the possible coordinates of point A?
\nAnswer:
\nIt is given that
\nPoint B has coordinates (2, 1). The x-coordinate of point A is -10. The distance between point A and point B is 15 units
\nNow,
\nLet the coordinates of A be:
\n(x, y) = (-10, y)
\nNow,
\nCompare the points A and B with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points A and B = \\(\\sqrt{(-10 – 2)^{2} + (y – 1)^{2}}\\)
\n15 = \\(\\sqrt{12^{2} + (y – 1)^{2}}\\)
\n15 = \\(\\sqrt{144 + (y – 1)^{2}}\\)
\nNow,
\nSquaring on both sides
\nSo,
\n144 + (y – 1)\u00b2 = 225
\n(y\u00a0 1)\u00b2 = 225 – 144
\n(y – 1)\u00b2 = 81
\ny – 1 = \\(\\sqrt{81}\\)
\ny – 1 = 9 (or) y – 1 = -9
\ny = 9 + 1 (or) y = -9 + 1
\ny = 10 (or) y = -8
\nHence, from the above,
\nWe can conclude that the possible coordinates of A are: (-10, 10), and (-10, -8)<\/p>\n

Question 13.
\nHigher-Order Thinking<\/strong> \u2206EFG and \u2206HIJ have the same perimeter and side lengths. The coordinates are E(6, 2), F(9, 2), G(8, 7), H(0, 0), and I(0, 3). What are the possible coordinates of point J?
\nAnswer:
\nIt is given that
\n\u2206EFG and \u2206HIJ have the same perimeter and side lengths. The coordinates are E(6, 2), F(9, 2), G(8, 7), H(0, 0), and I(0, 3)
\nSo,
\nAccording to the side lengths,
\nEF = HI, FG = IJ, and GE = JH
\nSo,
\nEF\u00b2 = HI\u00b2, FG\u00b2= JI\u00b2, and GE\u00b2 = JH\u00b2
\nNow,
\nLet the unknown vertex be J (x, y)
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points E and F = \\(\\sqrt{(9 – 6)^{2} + (2 – 2)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 0^{2}}\\)
\n= \\(\\sqrt{9 + 0}\\)
\n= 3 units
\nThe distance between the points F and G = \\(\\sqrt{(8 – 9)^{2} + (7 – 2)^{2}}\\)
\n= \\(\\sqrt{1^{2} + 5^{2}}\\)
\n= \\(\\sqrt{1 + 25}\\)
\n= 5.09 units
\nThe distance between the points G and E = \\(\\sqrt{(8 – 6)^{2} + (7 – 2)^{2}}\\)
\n= \\(\\sqrt{2^{2} + 5^{2}}\\)
\n= \\(\\sqrt{4 + 25}\\)
\n= 5.38 units
\nNow,
\nThe distance between the points H and I = \\(\\sqrt{(0 – 0)^{2} + (3 – 0)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 0^{2}}\\)
\n= \\(\\sqrt{9 + 0}\\)
\n= 3 units
\nThe distance between the points I and J = \\(\\sqrt{(x – 0)^{2} + (y – 3)^{2}}\\)
\n5.09 = \\(\\sqrt{x^{2} + (y – 3)^{2}}\\)
\nSquaring on both sides
\nSo,
\nx\u00b2 + (y – 3)\u00b2 = 25.90 units
\nThe distance between the points J and H = \\(\\sqrt{(x – 0)^{2} + (y – 0)^{2}}\\)
\n5.38 = \\(\\sqrt{x^{2} + y^{2}}\\)
\nSquaring on both sides
\nSo,
\nx\u00b2 + y\u00b2 = 28.94 units
\nNow,
\n28.94 – y\u00b2 + (y – 3)\u00b2 = 25.90
\ny\u00b2 + 9 – 6y – y\u00b2 = 25.90 – 28.94
\n9 – 6y = -3.04
\n-6y = -3.04 – 9
\n6y = 12.04
\ny = 2
\nNow,
\nSubstitute the value of y in eq 2
\nx\u00b2 + 4 = 28.94
\nx\u00b2 = 24.94
\nx = 4.99 (or) x = -4.99
\nHence, from the baove,
\nWe can conclude that the missing vertex is: J (4.99, 2) or J (-4.99, 2)<\/p>\n

b. Explain why there can be different possibilities for the coordinates for point J.
\nAnswer:<\/p>\n

Assessment Practice<\/strong>
\nQuestion 14.
\nFind the distance, in units, between P and R. Round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given coordinate plane is:
\n\"Envision
\nFrom the given coordinate plane,
\nThe points are:
\nP (5, 10), and R (12, 14)
\nNow,
\nCompare the points P and R with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points P and R = \\(\\sqrt{(14 – 10)^{2} + (12 – 5)^{2}}\\)
\n= \\(\\sqrt{4^{2} + 7^{2}}\\)
\n= \\(\\sqrt{16 + 49}\\)
\n= 8.1 units
\nHence, from the above,
\nWe can conclude that the distance between P and R is: 8.1 units<\/p>\n

Question 15.
\nFind the distance, in units, between A(1, 5) and B(5.5, 9.25). Round to the nearest tenth.
\nAnswer:
\nThe given points are:
\nA (1, 5), and B (5.5, 9.25)
\nNow,
\nCompare the points A and B with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points A and B = \\(\\sqrt{(9.25 – 5)^{2} + (5.5 – 1)^{2}}\\)
\n= \\(\\sqrt{4.25^{2} + 4.5^{2}}\\)
\n= \\(\\sqrt{18.06 + 20.25}\\)
\n= 6.2 units
\nHence, from the above,
\nWe can conclude that the distance between points A and B is: 6.2 units<\/p>\n

Topic 7 REVIEW<\/h3>\n

Topic Essential Question<\/strong>
\nHow can you use the Pythagorean Theorem to solve problems?
\nAnswer:
\nStep 1:
\nDraw a right triangle and then read through the problems again to determine the length of the legs and the hypotenuse. Step 2:
\nUse the Pythagorean Theorem\u00a0(a2<\/sup>\u00a0+ b2<\/sup>\u00a0= c2<\/sup>) to write an equation to be\u00a0solved.
\nStep 3:
\nSimplify the equation by distributing and combining like terms as needed.<\/p>\n

Vocabulary Review<\/strong><\/p>\n

Complete each definition and then provide an example of each vocabulary word.
\nVocabulary
\nThe converse of the Pythagorean Theorem
\nhypotenuse
\nleg
\nproof
\nPythagorean Theorem<\/p>\n

\"Envision
\nAnswer:
\n\"\"<\/p>\n

Use Vocabulary in Writing
\nAll faces of the figure are rectangles. Explain how to find the length of d. Use vocabulary terms in your description.
\n\"Envision
\nAnswer:
\nThe given figure is:
\n\"Envision
\nFrom the given figure,
\nWe can observe that there are 2 right triangles
\nNow,
\nFor the first right triangle,
\nThe given side lengths are:
\na = 3, b = 4, and c = x
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nSo,
\nx\u00b2 = 3\u00b2 + 4\u00b2
\nx\u00b2 = 9 + 16
\nx\u00b2 = 25
\nx = \\(\\sqrt{25}\\)
\nx = 5
\nNow,
\nFor the second right triangle,
\na = 5, b = 12, and c = d
\nSo,
\nd\u00b2 = 5\u00b2 + 12\u00b2
\nd\u00b2 = 25 + 144
\nd\u00b2 = 169
\nd = \\(\\sqrt{169}\\)
\nd = 13
\nHence, from the above,
\nWe can conclude that the length of d is: 13 units<\/p>\n

Concepts and Skills Review<\/strong><\/p>\n

Lesson 7.1 Understand the Pythagorean Theorem<\/strong><\/p>\n

Quick Review<\/strong>
\nThe Pythagorean Theorem states that, in a right triangle, the sum of the squares of the lengths of the legs, a and b, is equal to the square of the length of the hypotenuse, c. So, a2<\/sup> + b2<\/sup> = c2<\/sup>.<\/p>\n

Example<\/strong>
\nFind the length of the hypotenuse of a triangle with legs of 7 meters and 24 meters.
\nAnswer:
\nSubstitute 7 for a and 24 for b. Then solve for c.
\na2<\/sup> + b2<\/sup> = c2<\/sup>
\n49 + 576 = c2<\/sup>
\n\\(\\sqrt {625}\\) = C
\nThe length of the hypotenuse is 25 meters.<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nFind the length of the hypotenuse.
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the side lengths
\nSo,
\nc\u00b2 = 12\u00b2 + 5\u00b2
\nc\u00b2 = 144 + 25
\nc\u00b2 = 169
\nc = \\(\\sqrt{169}\\)
\nc = 13
\nHence, from the above,
\nWe can conclude that the length of the hypotenuse is: 13 cm<\/p>\n

Question 2.
\nFind the unknown side length. Round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given right triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2 + b\u00b2
\nWhere,
\nc is the hypotenuse
\na and b are the side lengths
\nSo,
\n9\u00b2 = 7\u00b2 + b\u00b2
\nb\u00b2 = 81 – 49
\nb\u00b2 = 32
\nb = \\(\\sqrt{32}\\)
\nb = 5.65
\nHence, from the above,
\nWe can conclude that the length of the hypotenuse is: 5.65 in.<\/p>\n

Lesson 7.2 Understand the Converse of the Pythagorean Theorem<\/strong><\/p>\n

Quick Review<\/strong>
\nFor a triangle with side lengths a, b, and c, if a2<\/sup> + b2<\/sup> = c2<\/sup>, then the triangle is a right triangle by the Converse of the Pythagorean Theorem.<\/p>\n

Example<\/strong>
\nIs a triangle with side lengths of 8 m, 15 m, and 17 m a right triangle? Explain.
\nAnswer:
\nSubstitute 8 for a, 15 for b, and 17 for c.
\na2<\/sup> + b2<\/sup> \\(\\underline{\\underline{?}}\\) c2<\/sup>
\n82<\/sup> + 152<\/sup> \\(\\underline{\\underline{?}}\\) 172<\/sup>
\n289 = 289 \u2713
\nBecause a2<\/sup> + b2<\/sup> = c2<\/sup>, the triangle is a right triangle.<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nIs the triangle a right triangle? Explain.
\n\"Envision
\nAnswer:
\nThe given triangle is:
\n\"Envision
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2
\nthen, the given triangle is a right triangle
\nSo,
\nNow,
\n37\u00b2 = 35\u00b2 + 12\u00b2
\n1,369 = 1,225 + 144
\n1,369 = 1,369
\nSo,
\nThe condition c\u00b2 = a\u00b2 + b\u00b2 is true
\nHence, from the above,
\nWe can conclude that the given triangle is a right triangle<\/p>\n

Question 2.
\nA triangle has side lengths of 1.5 inches, 2 inches, and 3 inches. Is the triangle a right triangle? Explain.
\nAnswer:
\nIt is given that
\nA triangle has side lengths of 1.5 inches, 2 inches, and 3 inches
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2
\nthen, the given triangle is a right triangle
\nWe know that,
\nThe longest side is the hypotenuse
\nSo,
\nNow,
\n3\u00b2 = (1.5)\u00b2 + 2\u00b2
\n9 = 2.25 + 4
\n9 = 6.25
\nSo,
\nThe condition c\u00b2 = a\u00b2 + b\u00b2 is false
\nHence, from the above,
\nWe can conclude that the given triangle is not a right triangle<\/p>\n

Question 3.
\nA triangle has side lengths of 9 feet, 40 feet, and 41 feet. Is the triangle a right triangle? Explain
\nAnswer:
\nIt is given that
\nA triangle has side lengths of 9 feet, 40 feet, and 41 feet
\nNow,
\nWe know that,
\nAccording to the converse of the Pythagorean Theorem,
\nIf
\nc\u00b2 = a\u00b2 + b\u00b2
\nthen, the given triangle is a right triangle
\nWe know that,
\nThe longest side is the hypotenuse
\nSo,
\nNow,
\n41\u00b2 = 40\u00b2 + 9\u00b2
\n1,681 = 1,600 + 81
\n1,681 = 1,681
\nSo,
\nThe condition c\u00b2 = a\u00b2 + b\u00b2 is true
\nHence, from the above,
\nWe can conclude that the given triangle is a right triangle<\/p>\n

Lesson 7.3 Apply the Pythagorean Theorem to Solve Problems<\/strong><\/p>\n

Quick Review<\/strong>
\nThe Pythagorean Theorem can be used to find unknown side lengths of an object that is shaped like a right triangle. It also can be used to find diagonal measures in certain two-dimensional and three-dimensional objects.<\/p>\n

Example<\/strong>
\nA shipping box is 20 inches long along the diagonal of its base. Each diagonal of the box is 29 inches long. How tall is the box?
\n\"Envision
\nAnswer:
\nSubstitute 20 for a and 29 for c. Then solve for b.
\na2<\/sup> + b2<\/sup> = c2<\/sup>
\n202<\/sup> + b2<\/sup> = 292<\/sup>
\n400 + b2<\/sup> = 841
\nb = \\(\\sqrt {441}\\)
\nThe height of the shipping box is 21 inches.<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nA basketball court is in the shape of a rectangle that is 94 feet long and 50 feet wide. What is the length of a diagonal of the court? Round to the nearest tenth.
\nAnswer:
\nIt is given that
\nA basketball court is in the shape of a rectangle that is 94 feet long and 50 feet wide
\nWe know that,
\nBy drawing a diagonal in the rectangle, it will become 2 right triangles
\nThe diagonal will be the hypotenuse of the right triangle
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the diagonal
\na and b are the side lengths
\nSo,
\nc\u00b2 = 94\u00b2+ 50\u00b2
\nc\u00b2= 8,836 + 2,500
\nc\u00b2 = 11,336
\nc = \\(\\sqrt{11,336}\\)
\nc = 106.4 feet
\nhence, from the above,
\nWe can conclude that the length of the diagonal is: 106.4 feet<\/p>\n

Question 2.
\nA packaging box for a metal rod is 7.5 inches along a diagonal of the base. The height of the box is 18 inches. What is the length of a diagonal of the box?
\n\"Envision
\nAnswer:
\nIt is given that
\nA packaging box for a metal rod is 7.5 inches along a diagonal of the base. The height of the box is 18 inches.
\nNow,
\nFrom the given figure,
\nWe can observe that it looks like a right triangle
\nNow,
\nWe know that,
\nAccording to the Pythagorean Theorem,
\nc\u00b2 = a\u00b2+ b\u00b2
\nWhere,
\nc is the length of the diagonal
\na and b are the side lengths
\nSo,
\nc\u00b2 = 18\u00b2 + (7.5)\u00b2
\nc\u00b2= 324 + 56.25
\nc\u00b2 = 380.25
\nc = \\(\\sqrt{380.25}\\)
\nc = 19.5 inches
\nHence, from the above,
\nWe can conclude that the length of the diagonal of the box is: 19.5 inches<\/p>\n

Lesson 7.4 Find Distance in the Coordinate Plane<\/strong><\/p>\n

Quick Review<\/strong>
\nThe Pythagorean Theorem can be used to find the distance between any two points on the coordinate plane.<\/p>\n

Example<\/strong>
\nFind the distance between the two points on the coordinate plane. Round to the nearest tenth.
\n\"Envision
\nAnswer:
\nDraw a right triangle. Determine the lengths of its legs.
\n\"Envision
\nThe length of the horizontal leg is 5 units.
\nThe length of the vertical leg is 5 units.
\nUse the relationship a2<\/sup> + b2<\/sup> = c2<\/sup>. Substitute 5 for a and 5 for b. Then solve for C.
\na2<\/sup> + b2<\/sup> = -2
\n52<\/sup> + 52<\/sup> = c2<\/sup>
\n25 + 25 = c2<\/sup>
\n50 = c2<\/sup>
\n\\(\\sqrt {50}\\) = C
\n7.1 \u2248 c
\nThe distance between the two points is about 7.1 units.<\/p>\n

Practice<\/strong>
\nQuestion 1.
\nPoints C and D represent the location of two parks on a map. Find the distance between the parks if the length of each unit on the grid is equal to 25 miles. Round to the nearest mile.
\n\"Envision
\nAnswer:
\nIt is given that
\nPoints C and D represent the location of two parks on a map
\nNow,
\nThe given coordinate plane is:
\n\"Envision
\nFrom the given coordinate plane,
\nThe points are:
\nC (-2, 2), and D (4, -1)
\nNow,
\nCompare the points C and D with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points C and D = \\(\\sqrt{(-1 – 2)^{2} + (4 + 2)^{2}}\\)
\n= \\(\\sqrt{3^{2} + 6^{2}}\\)
\n= \\(\\sqrt{9 + 36}\\)
\n= 6.70 units
\nIt is given that the length of each unit in the grid is: 25 miles
\nSo,
\nThe distance between the points C and D = 6.25 \u00d7 25
\n= 156.25 miles
\nHence, from the above,
\nWe can conclude that the distance between points C and D is: 156.25 miles<\/p>\n

Question 2.
\nFind the perimeter of \u2206ABC. Round to the nearest tenth.
\n\"Envision
\nAnswer:
\nThe given coordinate plane is:
\n\"Envision
\nNow,
\nFrom the given coordinate plane,
\nThe vertices of \u0394ABC are:
\nA (1, 2), B (7, 9), and C (7, 2)
\nWe know that,
\nThe perimeter of a triangle is the sum of all the side lengths of a triangle
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nFrom the given coordinate plane,
\nWe know that,
\nIn \u2206ABC,
\nAC and BC are the side lengths
\nAB is the hypotenuse
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points A and C = \\(\\sqrt{(2 – 2)^{2} + (7 – 1)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 6^{2}}\\)
\n= \\(\\sqrt{36 + 0}\\)
\n= 6 units
\nThe distance between the points B and C (BC) = \\(\\sqrt{(7 – 7)^{2} + (-2 + 9)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 7^{2}}\\)
\n= \\(\\sqrt{0 + 49}\\)
\n= 7 units
\nThe distance between the points A and B (AB) = \\(\\sqrt{(9 – 2)^{2} + (7 – 1)^{2}}\\)
\n= \\(\\sqrt{7^{2} + 6^{2}}\\)
\n= \\(\\sqrt{49 + 36}\\)
\n= 9.21 units
\nSo,
\nThe perimeter of \u2206ABC = AB + BC + AC
\n= 6 + 7 + 9.21
\n= 22.2 units
\nHence, from the above,
\nWe can conclude that the perimeter of \u2206ABC is about 22.2 units<\/p>\n

Question 3.
\nTriangle JKL is an equilateral triangle with two of its vertices at points J and K. What are the coordinates of point L? Round to the nearest tenth as needed.
\n\"Envision
\nAnswer:
\nIt is given that
\nTriangle JKL is an equilateral triangle with two of its vertices at points J and K
\nNow,
\nThe given coordinate plane is:
\n\"Envision
\nNow,
\nFrom the given coordinate plane,
\nThe two vertices of \u0394JKL are:
\nJ (3, 2), K (9, 2)
\nNow,
\nLet the third vertex be L (x, y)
\nWe know that,
\nIn an equilateral triangle, all the side lengths are the same
\nSo,
\nJK = KL = JL
\nJK\u00b2 = KL\u00b2 = JL\u00b2
\nSo,
\nNow,
\nCompare the given points with (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>)
\nNow,
\nWe know that,
\nThe distance between two points = \u221a(x2<\/sub>–\u00a0 x1<\/sub>) + (y2 <\/sub>– y1<\/sub>)\u00b2
\nSo,
\nThe distance between the points J and K = \\(\\sqrt{(9 – 3)^{2} + (2 – 2)^{2}}\\)
\n= \\(\\sqrt{0^{2} + 6^{2}}\\)
\n= \\(\\sqrt{36 + 0}\\)
\n= 6 units
\nThe distance between the points K and L = \\(\\sqrt{(x – 9)^{2} + (y – 2)^{2}}\\)
\n6= \\(\\sqrt{(x – 9)^{2} + (y – 2)^{2}}\\)
\nSquaring on both sides
\nSo,
\n(x – 9)\u00b2 + (y – 2)\u00b2 = 36
\nThe distance between the points J and L = \\(\\sqrt{(x – 3)^{2} + (y – 2)^{2}}\\)
\n6= \\(\\sqrt{(x – 3)^{2} + (y – 2)^{2}}\\)
\nSquaring on both sides
\nSo,
\n(x – 3)\u00b2 + (y – 2)\u00b2 = 36
\nNow,
\n(x – 9)\u00b2 + 36 – (x – 3)\u00b2 = 36
\n(x – 9)\u00b2 = (x – 3)\u00b2
\nx\u00b2 + 81 – 18x = x\u00b2 + 9 – 6x
\n18x – 6x = 81 – 9
\n12x = 72
\nx = \\(\\frac{72}{12}\\)
\nx = 6
\nSo,
\nSubstitute the value of x in eq (2)
\n(6 – 3)\u00b2 + (y – 2)\u00b2 = 36
\n(y – 2)\u00b2 = 36 – 9
\ny – 2 = \\(\\sqrt{27}\\)
\ny = 7.2
\nHence, from the above,
\nWe can conclude that the third vertex is: L (6, 7.2)<\/p>\n

Topic 7 Fluency Practice<\/h3>\n

Riddle Rearranging<\/strong>
\nSolve each equation. Then arrange the answers in order from least to greatest. The letters will spell out the answer to the riddle below.<\/p>\n

I can… solve multistep equations.
\n\"Envision<\/p>\n

Why did the coffee shop server love the job? Because there were so
\n\"Envision
\nAnswer:
\nThe Ascending order of the solutions of the equations is:
\nM < N < Y < P < A < E < R < K < S<\/p>\n","protected":false},"excerpt":{"rendered":"

Go through the\u00a0enVision Math Common Core Grade 8 Answer Key\u00a0Topic 7 Understand And Apply The Pythagorean Theorem\u00a0regularly and improve your accuracy in solving questions. enVision Math Common Core 8th Grade Answers Key Topic 7 Understand And Apply The Pythagorean Theorem Topic Essential Question How can you use the Pythagorean Theorem to solve problems? Answer: The\u00a0Pythagorean …<\/p>\n","protected":false},"author":1,"featured_media":25560,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/envisionmathanswerkey.com\/wp-content\/uploads\/2021\/09\/Envision-Math-Common-Core-Grade-8-Answer-Key-Topic-7-Understand-And-Apply-The-Pythagorean-Theorem.png","_links":{"self":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts\/4503"}],"collection":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/comments?post=4503"}],"version-history":[{"count":4,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts\/4503\/revisions"}],"predecessor-version":[{"id":27929,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/posts\/4503\/revisions\/27929"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/media\/25560"}],"wp:attachment":[{"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/media?parent=4503"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/categories?post=4503"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/envisionmathanswerkey.com\/wp-json\/wp\/v2\/tags?post=4503"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}