Trapezoid

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Trapezoid ABCD has bases AB and CD. Find x.

Guest Apr 24, 2022

CPhill · Answer 1 · 2022-04-24T08:47:02

Draw altitude AE

Then DE = (3 -1) / 2 = 1

And we have right triangle ADE with legs of sqrt 2 , 1 and the trapezoid height as the other leg

So

h = sqrt [( sqrt 2)^2 - 1 ] = 1

And we also have right triangle AEC where EC = 2

So

x = sqrt [ 2^2 + 1^2 ] = sqrt 5

BuilderBoi · Answer 2 · 2022-04-24T08:49:03

Draw line $AM$ so it is perpendicular to $DC$

Trapezoid $ABCD$ is isceles, meaning $DM = 1$

Applying the Pythagorean Theoram to $\triangle AMD$ , we find that $AM = 1$

Applying the Pythagorean Theorem to $\triangle AMC$ , we find $\color{brown}\boxed{x=\sqrt{5}}$