Question

What is the height of an isosceles trapezoid, if the lengths of its bases are 5m and 11m, and the length of a leg 4m.

Answer 1

Let EB and CF be two heights. Then quadrilateral EBCF is rectangle, so BC=EF=5 m. It is given that bases are BC=5 m and AD=11 m. Triangles ABE and DCF are congruent, thus,

$AE=FD=\dfrac{AD-EF}{2}=\dfrac{AD-BC}{2}=\dfrac{11-5}{2}=3.$

Consider right triangle ABE, by the Pythagorean theorem

$AB^2=EB^2+AE^2,\\ \\4^2=3^2+EB^2,\\ \\EB^2=16-9=7,\\ \\EB=\sqrt{7}\ m.$

Answer: $\sqrt{7}\ m.$