Find the height of the trapezoid

Geometry Level 1

A isosceles trapezoid has a base of length 12 12 and a top of length 4 4 . Find its height (length of the red line) if A C B D AC \perp BD .

Diagram not drawn to scale


The answer is 8.

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3 solutions

Chaitanya Rao
Dec 23, 2020

If h h is the height, set coordinates as A ( 4 , h ) , B ( 0 , 0 ) , C ( 12 , 0 ) , D ( 8 , h ) A(4,h), B(0,0), C(12,0), D(8,h) . Since A C B D AC\perp BD , the vectors ( 8 , h ) (8,-h) and ( 8 , h ) (8,h) are perpedicular, so their inner product 64 h 2 64 - h^2 is 0 0 . As h > 0 h > 0 this implies h = 8 \boxed{h = 8} .

David Vreken
Dec 7, 2020

Draw E M A D EM \perp AD and F M B C FM \perp BC .

Then M E A M E D \triangle MEA \cong \triangle MED by HL congruence, and A M E = D M E = 45 ° \angle AME = \angle DME = 45° , so M E A \triangle MEA is an isosceles right triangle so that E M = A E = 2 EM = AE = 2 .

Similarly, M F B M F C \triangle MFB \cong \triangle MFC by HL congruence, and B M F = C M F = 45 ° \angle BMF = \angle CMF = 45° , so M F B \triangle MFB is an isosceles right triangle so that F M = B F = 6 FM = BF = 6 .

Therefore, the height of the trapezoid is E M + F M = 2 + 6 = 8 EM + FM = 2 + 6 = \boxed{8} .

Toby M
Dec 6, 2020

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 = 90 º \angle AMD = \angle BMC = 90º , by Pythagoras B M 2 + C M 2 = 1 2 2 BM^2 + CM^2 = 12^2 , and since the trapezoid is isosceles, B M = C M BM = CM . Therefore, 2 B M 2 = 1 2 2 B M = 6 2 2 BM^2 = 12^2 \Rightarrow BM = 6 \sqrt{2} . Similarly, A M = 2 2 AM = 2 \sqrt{2} .

Since A M B AMB is a right angle as well, A B 2 = ( 2 2 ) 2 + ( 6 2 ) 2 = 80 AB^2 = (2 \sqrt{2})^2 + (6 \sqrt{2})^2 = 80 . Therefore B N 2 + A N 2 = A B 2 BN^2 + AN^2 = AB^2 and B N = 1 2 ( 12 4 ) = 4 BN = \frac{1}{2}(12 - 4) = 4 , so the red line A N = 80 4 2 = 8 AN = \sqrt{80 - 4^2} = 8 .

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