What is the perimeter of the trapezoid above?

52

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72

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75

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80

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87

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Solution

The correct option is B $80$
In given figure, $A B = 20, B C = 15$ and $A D = 17$
Draw perpendicular from point $A$ to line $C D$ meet at point $E$
Then $B C = A E = 15$
So in $△ A E D$ ,
$D E =$ $\sqrt{(17)^{2} - (15)^{2}} = \sqrt{289 - 225} = \sqrt{64} = 8$
Then $D C = D E + C E = 8 + 20 = 28$ .......( $A B = C E = 20$ )
Then perimeter of trapezoid $= 20 + 17 + 15 + 28 = 80$

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What is the perimeter of the trapezoid above?

The correct option is B 80In given figure, AB=20,BC=15 and AD=17Draw perpendicular from point A to line CD meet at point E Then BC=AE=15So in △AED,DE= √(17)2−(15)2=√289−225=√64=8Then DC=DE+CE=8+20=28 .......( AB=CE=20)Then perimeter of trapezoid =20+17+15+28=80