Chapter 11.2, Problem 39STP

To determine

To calculate: The area of given figure.

Answer to Problem 39STP

Area is (A) $162{\text{ m}}^2$ .

Explanation of Solution

Given information: Lengths of the bases of an isosceles trapezoid are 19 m and 35 m.

Perimeter of given isosceles trapezoid is 74 m.

Formula used:Area of isosceles trapezoid is given by

  $\text{Area of isosceles trapezoid}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}$

Calculation: In isosceles trapezoid, two opposite sides(the bases) are parallel, and the two other sides(the legs) are of equal length.

Let the remaining unknown sides of isosceles trapezoid be x .

  $\begin{array}{l}\text{Perimeter of rhombus}=(\text{Sum of two parallel sides)}+2x\\ \Rightarrow 74=(19+35)+2x\\ \Rightarrow 74=54+2x\\ \Rightarrow 2x=20\\ \Rightarrow x=10\text{ m}\end{array}$

Draw perpendiculars from 19 m side to 35 m side.

  

Then, by Pythagoras theorem,

  $\begin{array}{l}\text{Height of perpendicular}=\sqrt{{(10)}^2-{(8)}^2}\\ \text{ =}\sqrt{100-64}\\ =\sqrt{36}\\ =6\text{ m}\end{array}$

  $\begin{array}{l}\text{Area of isosceles trapezoid}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}\\ =\frac{1}{2}\times (19+35)\times 6\\ =54\times 3\\ =162{\text{ m}}^2\end{array}$

Thus, $\text{Area of isosceles trapezoid}=162{\text{ m}}^2$ .

Hence, (A) $162{\text{ m}}^2$ is the area of the given isosceles trapezoid.