Chapter 11.3, Problem 57SPR

To determine

To find: ‘Area of Parallelogram ABCD’

Answer to Problem 57SPR

Area of Parallelogram ABCD = 178.2 sq.inches

Explanation of Solution

Given Information:

  $\begin{array}{l}AD=BC=9\\ DC=AB=28\\ \angle D=45°\\ \angle DMA=90°\end{array}$

Figure:

  

Solution:

  $\begin{array}{l}\text{Area Of Parallelogram = Base }\times \text{ Height}\\ \text{Ar}\parallel ABCD=DC\times AM\end{array}$

Here,

DC = Base &

AM = Height of the Parallelogram

Now,

Since height of the parallelogram is not given, find it out by Sine formula using the Trigonometric ratios.

In Δ ADL,

  $\begin{array}{l}\text{Sinθ = }\frac{\text{Perpendicular Height}}{\text{Hypotenuse}}\\ \Rightarrow \text{SinD = }\frac{\text{AM}}{\text{AD}}\\ \Rightarrow \text{Sin45° = }\frac{\text{AM}}{\text{9}}\\ \Rightarrow \text{AM = Sin45°×9}\\ \Rightarrow \text{AM}=\frac{\sqrt{2}}{2}\times 9\\ \Rightarrow \text{AM}=\frac{12.726}{2}\\ \Rightarrow \text{AM}=6.363\text{ ft}\text{.}\end{array}$

Thus,

  $\begin{array}{l}\text{Area of Parallelogram ABCD = DC }\times \text{ AM}\\ \text{Ar}\parallel ABCD=28\times 6.363\\ \text{Ar}\parallel ABCD=178.164\text{ sq}\text{.inches}\\ \text{Ar}\parallel ABCD\approx 178.2\text{ sq}\text{.inches}\end{array}$

Therefore,

Area of Parallelogram ABCD = 178.2 sq.inches