Chapter 12.3, Problem 55SR

To determine

To find:

The perimeter and area of the parallelogram.

Answer to Problem 55SR

The perimeter is $57\text{ m}$ and area is $120{\text{ mm}}^2$ .

Explanation of Solution

Given information:

  

Calculation:

From the given figure.

The triangle inside the parallelogram is ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }$ .

So, the height of the triangle is $6\text{ m}$ .

Thus, the hypotenuse is as follows:

  $\begin{array}{c}h=\sqrt{6^2+6^2}\\ =6\sqrt{2}\end{array}$

Therefore, the perimeter of the parallelogram is:

  $\begin{array}{c}P=20+20+6\sqrt{2}+6\sqrt{2}\\ =40+12\sqrt{2}\\ \approx 57\text{ m}\end{array}$

Area of the parallelogram is given as:

  $\begin{array}{c}A=bh\\ =20\cdot 6\\ =120{\text{ mm}}^2\end{array}$

Hence, the perimeter is $57\text{ m}$ and area is $120{\text{ mm}}^2$ .