Chapter 11.3, Problem 58SPR

To determine

To find: ‘Area of the Parallelogram ABCD’

Answer to Problem 58SPR

Area of Parallelogram ABCD = 315.2 sq.cm

Explanation of Solution

Given Information:

  $\begin{array}{l}AD=BC=26\\ DC=AB=14\\ \angle A=30°\\ \angle DTA=90°\end{array}$

Figure:

  

Solution:

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

Here,

DC = Base &

AT = Height of the Parallelogram

Now,

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

In Δ ADL,

  $\begin{array}{l}\text{Cosθ = }\frac{\text{Base/Adjacent Side}}{\text{Hypotenuse}}\\ \Rightarrow \text{CosA = }\frac{\text{AT}}{\text{AD}}\\ \Rightarrow \text{Cos30° = }\frac{\text{AT}}{\text{26}}\\ \Rightarrow \text{AT = Cos30°×26}\\ \Rightarrow \text{AT}=\frac{\sqrt{3}}{2}\times 26\\ \Rightarrow \text{AT}=\sqrt{3}\times 13\\ \Rightarrow \text{AT}=22.516\text{ cm}\end{array}$

Thus,

  $\begin{array}{l}\text{Area of Parallelogram ABCD = DC }\times \text{ AT}\\ \text{Ar}\parallel ABCD=14\times 22.52\\ \text{Ar}\parallel ABCD=315.224\text{ sq}\text{.cm}\\ \text{Ar}\parallel ABCD\approx 315.2\text{ sq}\text{.cm}\end{array}$

Therefore,

Area of Parallelogram ABCD = 315.2 sq.cm