{"id":82614,"date":"2022-06-02T18:29:24","date_gmt":"2022-06-02T12:59:24","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=82614"},"modified":"2022-05-10T14:19:21","modified_gmt":"2022-05-10T08:49:21","slug":"eureka-math-grade-8-module-2-lesson-16","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-2-lesson-16\/","title":{"rendered":"Eureka Math Grade 8 Module 2 Lesson 16 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 2 Lesson 16 Answer Key<\/h2>\n

Engage NY Math 8th Grade Module 2 Lesson 16 Example Answer Key<\/h3>\n

Example 1.
\nGiven a right triangle with a hypotenuse with length 13 units and a leg with length 5 units, as shown, determine the length of the other leg.
\n\"Engage
\nAnswer:
\n52<\/sup> +b2<\/sup> =132<\/sup>
\n52<\/sup> – 52<\/sup> + b2<\/sup> =132<\/sup> -52<\/sup>
\nb2<\/sup> =132<\/sup> -52<\/sup>
\nb2<\/sup> =169-25
\nb2<\/sup> =144
\nb=12
\nThe length of the leg is 12 units.<\/p>\n

\u2192 Let b represent the missing leg of the right triangle; then, by the Pythagorean theorem:
\n52<\/sup> +b2<\/sup> =132<\/sup>.
\n\u2192 If we let a represent the missing leg of the right triangle, then by the Pythagorean theorem:
\na2<\/sup> +52<\/sup> =132<\/sup>.
\n\u2192 Which of these two equations is correct: 52<\/sup> +b2<\/sup> =132<\/sup> or a2<\/sup> +52<\/sup> =132<\/sup> ?<\/p>\n

\u2192 It does not matter which equation we use as long as we are showing the sum of the squares of the legs as equal to the square of the hypotenuse.
\n\u2192 Using the first of our two equations, 52<\/sup> +b2<\/sup> =132<\/sup>, what can we do to solve for b in the equation?
\n\u2192 We need to subtract 52<\/sup> from both sides of the equation.
\n52<\/sup> + b2<\/sup> =132<\/sup>
\n52<\/sup> – 52<\/sup> + b2<\/sup> =132<\/sup> -52<\/sup>
\nb2<\/sup> =132<\/sup> -52<\/sup>
\n\u2192 Point out to students that we are looking at the Pythagorean theorem in a form that allows us to find the length of one of the legs of the right triangle. That is, b2<\/sup> =c2<\/sup> -a2<\/sup> .
\n\u2192 The length of the leg of the right triangle is 12 units.<\/p>\n

Example 2.
\nThe Pythagorean theorem as it applies to missing side lengths of triangles in a real-world problem:
\n\u2192 Suppose you have a ladder of length 13 feet. Suppose that to make it sturdy enough to climb, you must place the ladder exactly 5 feet from the wall of a building. You need to post a banner on the building 10 feet above the ground. Is the ladder long enough for you to reach the location you need to post the banner?
\n\"Engage
\nThe ladder against the wall forms a right angle. For that reason, we can use the Pythagorean theorem to find out how far up the wall the ladder will reach. If we let h represent the height the ladder can reach, what equation will represent this problem?
\n\u2192 52<\/sup> +h2<\/sup> =132<\/sup> or h2<\/sup> =132<\/sup> -52<\/sup><\/p>\n

\u2192 Using either equation, we see that this is just like Example 1. We know that the missing side of the triangle is 12 feet. Is the ladder long enough for you to reach the 10-foot banner location?
\n\u2192 Yes, the ladder allows us to reach 12 feet up the wall.<\/p>\n

Example 3.
\nPythagorean theorem as it applies to missing side lengths of a right triangle:
\n\u2192 Given a right triangle with a hypotenuse of length 15 units and a leg of length 9, what is the length of the other leg?
\n\"Engage
\n\u2192 If we let the length of the missing leg be represented by a, what equation will allow us to determine its value?
\n\u2192 a2<\/sup> +92<\/sup> =152<\/sup> or a2<\/sup> =152<\/sup> -92<\/sup>
\n\u2192 Finish the computation:
\na2<\/sup> =225-81
\na2<\/sup> =144
\na=12
\n\u2192 The length of the missing leg of this triangle is 12 units.<\/p>\n

Eureka Math Grade 8 Module 2 Lesson 16 Exercise Answer Key<\/h3>\n

Exercises 1\u20132<\/p>\n

Exercise 1.
\nUse the Pythagorean theorem to find the missing length of the leg in the right triangle.
\nAnswer:
\n\"Eureka
\nLet b represent the missing leg length; then,
\n152<\/sup> +b2<\/sup> =252<\/sup>
\n152<\/sup> -152<\/sup> +b2<\/sup> =252<\/sup> -152<\/sup>
\nb2<\/sup> =625-225
\nb2<\/sup> =400
\nb=20
\nThe length of the leg is 20 units.<\/p>\n

Exercise 2.
\nYou have a 15-foot ladder and need to reach exactly 9 feet up the wall. How far away from the wall should you place the ladder so that you can reach your desired location?
\nAnswer:
\n\"Eureka
\nLet a represent the distance the ladder must be placed from the wall; then,
\na2<\/sup> +92<\/sup> =152<\/sup>
\na2<\/sup> +92<\/sup> -92<\/sup> =152<\/sup> -92<\/sup>
\na2<\/sup> =225-81
\na2<\/sup> =144
\na=12.
\nThe ladder must be placed exactly 12 feet from the wall.<\/p>\n

Exercises 3\u20136<\/p>\n

Exercise 3.
\nFind the length of the segment AB, if possible.
\n\"Eureka
\nAnswer:
\nIf we let the length of segment AB be represented by c, then
\n32<\/sup> +42<\/sup> =c2<\/sup>
\n9+16=c2<\/sup>
\n25=c2<\/sup>
\n5=c.
\nThe length of segment AB is 5 units.
\n\"Eureka<\/p>\n

Exercise 4.
\nGiven a rectangle with dimensions 5 cm and 10 cm, as shown, find the length of the diagonal, if possible.
\n\"Eureka
\nAnswer:
\nLet c represent the length of the diagonal; then,
\nc2<\/sup> =52<\/sup> +102<\/sup>
\nc2<\/sup> =25+100
\nc2<\/sup> =125.
\nThe measure of the length of the hypotenuse in centimeters is the positive number c that satisfies c2<\/sup> =125.<\/p>\n

Exercise 5.
\nA right triangle has a hypotenuse of length 13 in. and a leg with length 4 in. What is the length of the other leg?
\nAnswer:
\nIf we let a represent the length of the other leg, then
\na2<\/sup> +42<\/sup> =132<\/sup>
\na2<\/sup> +42<\/sup> -42<\/sup> =132<\/sup> -42<\/sup>
\na2<\/sup> =132<\/sup> -42<\/sup>
\na2<\/sup> =169-16
\na2<\/sup> =153
\nThe measure of the length of the leg in inches is the positive number a that satisfies a2<\/sup> =153.<\/p>\n

Exercise 6.
\nFind the length of b in the right triangle below, if possible.
\n\"Eureka
\nAnswer:
\nBy the Pythagorean theorem,
\n42<\/sup> +b2<\/sup> =112<\/sup>
\n42<\/sup> -42<\/sup> +b2<\/sup> =112<\/sup> -42<\/sup>
\nb2<\/sup> =112<\/sup> -42<\/sup>
\nb2<\/sup> =121-16
\na2<\/sup> =105.
\nThe length of side AC is the positive number b that satisfies b2<\/sup> =105.<\/p>\n

Eureka Math Grade 8 Module 2 Lesson 16 Problem Set Answer Key<\/h3>\n

Students practice using the Pythagorean theorem to find missing lengths in right triangles.<\/p>\n

Question 1.
\nFind the length of the segment AB shown below, if possible.
\n\"Eureka
\nAnswer:
\nIf we let the length of segment AB be represented by c units, then by the Pythagorean theorem
\n62<\/sup> +82<\/sup> =c2<\/sup>
\n36+48=c2<\/sup>
\n100=c2<\/sup>
\n10=c
\nThe length of the segment AB is 10 units.
\n\"Eureka<\/p>\n

Question 2.
\nA 20-foot ladder is placed 12 feet from the wall, as shown. How high up the wall will the ladder reach?
\n\"Eureka
\nAnswer:
\nLet the height up the wall that the ladder will reach be a feet. Then,
\na2<\/sup> +122<\/sup> =202<\/sup>
\na2<\/sup> +122<\/sup> -122<\/sup> =202<\/sup> -122<\/sup>
\na2<\/sup> =202<\/sup> -122<\/sup>
\na2<\/sup> =400-144
\na2<\/sup> =256
\na=16
\nThe ladder will reach 16 feet up the wall.<\/p>\n

Question 3.
\nA rectangle has dimensions 6 in. by 12 in. What is the length of the diagonal of the rectangle?
\nAnswer:
\nLet the length of the diagonal be c inches. Then c is a positive number that satisfies
\n62<\/sup> +122<\/sup> =c2<\/sup>
\n36+144=c2<\/sup>
\n80=c2<\/sup> .
\nThe measure of the length of the diagonal in inches is the positive number c that satisfies c2<\/sup> =180.<\/p>\n

Use the Pythagorean theorem to find the missing side lengths for the triangles shown in Problems 4\u20138.<\/p>\n

Question 4.
\nDetermine the length of the missing side, if possible.
\n\"Eureka
\n122<\/sup>+b2<\/sup>=132<\/sup>
\n122<\/sup>-122<\/sup>+b2<\/sup>=\u3016132<\/sup>-122<\/sup>
\nb2<\/sup> =132<\/sup>-122<\/sup>
\nb2<\/sup>=169-144
\nb2<\/sup>=25
\nb=5
\nThe length of the missing side is 5 units.<\/p>\n

Question 5.
\nDetermine the length of the missing side, if possible.
\n\"Eureka
\nAnswer:
\na2<\/sup>+32<\/sup>=82<\/sup>
\na2<\/sup>+32<\/sup>-32<\/sup>=82<\/sup>-32<\/sup>
\na2<\/sup>=82<\/sup>-32<\/sup>
\na2<\/sup>=64-9
\na2<\/sup>=55)
\nThe number of units of the side is given by the positive number a that satisfies a2<\/sup>=55.<\/p>\n

Question 6.
\nDetermine the length of the missing side, if possible.
\n\"Eureka
\nAnswer:
\n72<\/sup>+b2<\/sup>=102<\/sup>
\n72<\/sup>-72<\/sup>+b2<\/sup>=102<\/sup>-72<\/sup>
\nb2<\/sup>=102<\/sup>-72<\/sup>
\nb2<\/sup>=100-49
\nb2<\/sup>=51
\nThe number of units of the side is given by the positive number b that satisfies b2<\/sup>=51.<\/p>\n

Question 7.
\nDetermine the length of the missing side, if possible.
\n\"Eureka
\nAnswer:
\na2<\/sup>+12<\/sup>=52<\/sup>
\na2<\/sup>+12<\/sup>-12<\/sup>=52<\/sup>-12<\/sup>
\na2<\/sup>=52<\/sup>-12<\/sup>
\na2<\/sup>=25-1
\na2<\/sup>=24
\nThe number of units of the side is given by the positive number a that satisfies a2<\/sup>=24.<\/p>\n

Question 8.
\nDetermine the length of the missing side, if possible.
\n\"Eureka
\nAnswer:
\na2<\/sup>+92<\/sup>=142<\/sup>
\na2<\/sup>+92<\/sup>-92<\/sup>=142<\/sup>-92<\/sup>
\na2<\/sup>=142<\/sup>-92<\/sup>
\na2<\/sup>=196-81
\na2<\/sup>=115
\nThe number of units of the side is given by the positive number a that satisfies a2<\/sup>=115.<\/p>\n

Eureka Math Grade 8 Module 2 Lesson 16 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nFind the length of the missing side of the rectangle shown below, if possible.
\n\"Eureka
\nAnswer:
\nLet the length of the unknown leg be a units. Then,
\na2<\/sup> +72<\/sup> =122<\/sup>
\na2<\/sup> +72<\/sup> -72<\/sup> =122<\/sup> -72<\/sup>
\na2<\/sup> =122<\/sup> -72<\/sup>
\na2<\/sup> =144-49
\na2<\/sup> =95.
\nThe number of units of the side is given by the positive number a that satisfies a2<\/sup> =95.<\/p>\n

Question 2.
\nFind the length of all three sides of the right triangle shown below, if possible.
\n\"Eureka
\nAnswer:
\n\"Eureka
\nThe two legs are each 5 units in length. If the hypotenuse is c units in length, then c is a positive number that satisfies
\n52<\/sup>+52<\/sup>=c2<\/sup>
\n25+25=c2<\/sup>
\n50=c2<\/sup>.
\nThe length of the hypotenuse is given by the positive number c that satisfies c2<\/sup> =50 units.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 2 Lesson 16 Answer Key Engage NY Math 8th Grade Module 2 Lesson 16 Example Answer Key Example 1. Given a right triangle with a hypotenuse with length 13 units and a leg with length 5 units, as shown, determine the length of the other leg. Answer: 52 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82614"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=82614"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82614\/revisions"}],"predecessor-version":[{"id":137314,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82614\/revisions\/137314"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=82614"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=82614"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=82614"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}