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Arch. The stone above the arch in the diagram is an isosceles trapezoid. Find m K , m M , and m J. STEP 1. Find m K . JKLM is an isosceles trapezoid, so K and L are congruent base angles, and m K = m L = 85 o. EXAMPLE 2. Use properties of isosceles trapezoids. SOLUTION.
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Arch The stone above the arch in the diagram is an isosceles trapezoid. Find mK, mM, and mJ. STEP 1 Find mK. JKLMis an isosceles trapezoid, so Kand Lare congruent base angles, and mK = mL= 85o. EXAMPLE 2 Use properties of isosceles trapezoids SOLUTION
Find m M. Because Land M are consecutive interior angles formed by LMintersecting two parallel lines, they are supplementary. So, mM = 180o – 85o = 95o. STEP 3 Find mJ. Because J and M are a pair of base angles, they are congruent, and mJ = mM= 95o. ANSWER So, mJ = 95o, m K = 85o, and m M = 95o. EXAMPLE 2 Use properties of isosceles trapezoids STEP 2
In the diagram,MNis the midsegment of trapezoidPQRS. FindMN. SOLUTION Use Theorem 8.17 to findMN. 1 =(12+ 28) 2 1 MN (PQ + SR) = 2 ANSWER The length MNis 20 inches. EXAMPLE 3 Use the midsegment of a trapezoid Apply Theorem 8.17. Substitute 12 for PQand 28 for XU. = 20 Simplify.
3. If EG = FH, is trapezoid EFGHisosceles? Explain. ANSWER yes, Theorem 8.16 for Examples 2 and 3 GUIDED PRACTICE In Exercises 3 and 4, use the diagram of trapezoidEFGH.
4. If mHEF = 70o and mFGH =110o, is trapezoid EFGHisosceles? Explain. SAMPLE ANSWER Yes; mEFG =70° by Consecutive Interior Angles Theorem making EFGH an isosceles trapezoidby Theorem 8.15. for Examples 2 and 3 GUIDED PRACTICE
J K 9 cm ANSWER 12 cm P N M L 1 2 15 cm; Solve for x to find ML. ( 9 + x ) = 12 for Examples 2 and 3 GUIDED PRACTICE 5. In trapezoid JKLM, Jand M are right angles, and JK = 9 cm. The length of the midsegment NPof trapezoid JKLMis 12 cm. Sketch trapezoid JKLMand its midsegment. Find ML. Explain your reasoning.