Question

what is the height of an isosceles trapezoid if the lengths of its bases are 5 m and 11 m and the length of a leg is 4 m? answer: m

Explanation:

Step1: Calculate base - length difference

Let the lengths of the bases be $a = 5$ m and $b = 11$ m. The difference between the two bases is $b - a=11 - 5 = 6$ m. In an isosceles trapezoid, the non - parallel sides are equal. If we consider the right - triangle formed by the height, part of the base difference, and the non - parallel side (leg), the base of this right - triangle is $\frac{b - a}{2}=\frac{6}{2}=3$ m.

Step2: Use Pythagorean theorem

The length of the leg of the isosceles trapezoid (hypotenuse of the right - triangle) is $l = 4$ m, and the base of the right - triangle is $x = 3$ m. Let the height of the isosceles trapezoid be $h$. According to the Pythagorean theorem $h=\sqrt{l^{2}-x^{2}}$. Substitute $l = 4$ m and $x = 3$ m into the formula: $h=\sqrt{4^{2}-3^{2}}=\sqrt{16 - 9}=\sqrt{7}$ m.

Answer:

$\sqrt{7}$