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EXAMPLE 2 Use properties of isosceles trapezoids Arch The stone above the arch in the diagram is an isosceles trapezoid. Find m K, m M, and m J. SOLUTION.

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Presentation on theme: "EXAMPLE 2 Use properties of isosceles trapezoids Arch The stone above the arch in the diagram is an isosceles trapezoid. Find m K, m M, and m J. SOLUTION."— Presentation transcript:

1 EXAMPLE 2 Use properties of isosceles trapezoids Arch The stone above the arch in the diagram is an isosceles trapezoid. Find m K, m M, and m J. SOLUTION 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.

2 EXAMPLE 2 Use properties of isosceles trapezoids STEP 2 Find m M. Because L and M are consecutive interior angles formed by LM intersecting two parallel lines, they are supplementary. So, m M = 180 o – 85 o = 95 o. STEP 3 Find m J. Because J and M are a pair of base angles, they are congruent, and m J = m M = 95 o. ANSWER So, m J = 95 o, m K = 85 o, and m M = 95 o.

3 EXAMPLE 3 Use the midsegment of a trapezoid SOLUTION Use Theorem 8.17 to find MN. In the diagram, MN is the midsegment of trapezoid PQRS. Find MN. MN (PQ + SR) 1 2 = Apply Theorem 8.17. = (12 + 28) 1 2 Substitute 12 for PQ and 28 for XU. Simplify. = 20 ANSWERThe length MN is 20 inches.

4 GUIDED PRACTICE for Examples 2 and 3 In Exercises 3 and 4, use the diagram of trapezoid EFGH. 3. If EG = FH, is trapezoid EFGH isosceles? Explain. ANSWERyes, Theorem 8.16

5 GUIDED PRACTICE for Examples 2 and 3 4. If m HEF = 70 o and m FGH = 110 o, is trapezoid EFGH isosceles? Explain. SAMPLE ANSWERYes; m EFG = 70° by Consecutive Interior Angles Theorem making EFGH an isosceles trapezoid by Theorem 8.15.

6 GUIDED PRACTICE for Examples 2 and 3 5. In trapezoid JKLM, J and M are right angles, and JK = 9 cm. The length of the midsegment NP of trapezoid JKLM is 12 cm. Sketch trapezoid JKLM and its midsegment. Find ML. Explain your reasoning. J L K M 9 cm 12 cm N P ANSWER ( 9 + x ) = 12 1 2 15 cm; Solve for x to find ML.


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