Chapter 11, Problem 6RP

To determine

To calculate:Thearea of anisosceles trapezoid ABCD .

Answer to Problem 6RP

The area of an isosceles trapezoid ABCDis $288$ .

Explanation of Solution

Given information:

In anisosceles trapezoidABCD ,

Side AB = $20$ ,

Side AD = $8$ ,

Side DC = $20$ ,

Side BC = $40$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of trapezoid: $A=\frac{1}{2}\times (b_1+b_2)\times h$

b = base of trapezoid

h = height of trapezoid

Calculation:

  

To find h , draw two altitudes of the trapezoid.The middle figure formed is rectangle.

x = 8

y and z are congruent because it is an isosceles trapezoid.

  $\begin{array}{l}40-x=y+z\\ 40-8=y+z\\ 32=y+z\\ y\cong z\\ 32=2y\\ y=\frac{1}{2}\times 32=16\end{array}$

The value of h can be calculated by applying Pythagoras Theorem.

In right angle triangle AEB , we get

  ${\left(AB\right)}^2={\left(AE\right)}^2+{\left(BE\right)}^2$

  $\begin{array}{l}{\left(20\right)}^2={\left(h\right)}^2+{\left(16\right)}^2\\ {\left(h\right)}^2=400-256\\ {\left(h\right)}^2=144\\ h=12\end{array}$

Area of trapezoid ABCD :

  $\begin{array}{l}A=\frac{1}{2}\times (BC+AD)\times h\\ A=\frac{1}{2}\times (40+8)\times 12\\ A=48\times 6\\ A=288\end{array}$