A isosceles trapezoid has a base of length 1 2 and a top of length 4 . Find its height (length of the red line) if A C ⊥ B D .
Diagram not drawn to scale
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Draw E M ⊥ A D and F M ⊥ B C .
Then △ M E A ≅ △ M E D by HL congruence, and ∠ A M E = ∠ D M E = 4 5 ° , so △ M E A is an isosceles right triangle so that E M = A E = 2 .
Similarly, △ M F B ≅ △ M F C by HL congruence, and ∠ B M F = ∠ C M F = 4 5 ° , so △ M F B is an isosceles right triangle so that F M = B F = 6 .
Therefore, the height of the trapezoid is E M + F M = 2 + 6 = 8 .
Let the intersection of AC and BD be M, and the foot of the perpendicular on BC be N. Then since ∠ A M D = ∠ B M C = 9 0 º , by Pythagoras B M 2 + C M 2 = 1 2 2 , and since the trapezoid is isosceles, B M = C M . Therefore, 2 B M 2 = 1 2 2 ⇒ B M = 6 2 . Similarly, A M = 2 2 .
Since A M B is a right angle as well, A B 2 = ( 2 2 ) 2 + ( 6 2 ) 2 = 8 0 . Therefore B N 2 + A N 2 = A B 2 and B N = 2 1 ( 1 2 − 4 ) = 4 , so the red line A N = 8 0 − 4 2 = 8 .
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If h is the height, set coordinates as A ( 4 , h ) , B ( 0 , 0 ) , C ( 1 2 , 0 ) , D ( 8 , h ) . Since A C ⊥ B D , the vectors ( 8 , − h ) and ( 8 , h ) are perpedicular, so their inner product 6 4 − h 2 is 0 . As h > 0 this implies h = 8 .