Chapter 11, Problem 2PT

To determine

To Compute : The area and perimeter of the given figure.

Answer to Problem 2PT

The perimeter and area of the given figure are 92.5 in and 500 in²respectively.

Explanation of Solution

Given information :

  

Using Pythagoras theorem,

  $\text{AC = }\sqrt{A{D}^2+C{D}^2}$

  $\text{AC}=\sqrt{{20}^2+2^2}=\sqrt{404}$

And,

  $\text{BF = }\sqrt{B{E}^2+E{F}^2}$

  $\text{BF}=\sqrt{{20}^2+{10}^2}=\sqrt{500}$ .

Therefore,

Perimeter of (ABFC) = Sum of lengths of all the sides of (ABFC)

= Length of (AB + BF + FC + AC)

= $19+\sqrt{500}\text{ + 31 + }\sqrt{404}$

  $=92.5in$

Area of (ABFC) = Area of ( $△ACD$) + Area of rectangle (ABED) + Area of ( $△BEF$)

As, $\text{Area of a triangle = }\frac{1}{2}\times \text{base×height}$,

Implies,

Area of ( $△ACD$) = $\frac{1}{2}\times 2\times 20=20i{n}^2$

Area of ( $△BEF$) = $\frac{1}{2}\times 10\times 20=100i{n}^2$

And as,

  $\text{Area of a rectangle = length×breadth}$

Implies,

Area of rectangle (ABED) = $\text{20}\times \text{19 = 380 }i{n}^2$.

Therefore,

Area of (ABFC) = 20 + 100 + 380 = 500 in².

Thus,

The perimeter and area of the given figure are 92.5 in and 500 in²respectively.