Chapter 11.4, Problem 4ACYP

To determine

To calculate: The area of the given figure.

Answer to Problem 4ACYP

The area of the give figure is $148.5{\text{ in}}^2$.

Explanation of Solution

Given information: The length of four sides of the figure is $5\text{in,8in,12in and 15in}$ .

Formula Used: The area of the rectangle is,

  $A_R=L\times B$......(1)

• Here, $A_R$ is the area of the rectangle, L is the length and B is the breadth of the rectangle.

The area of the triangle is,

  $A_T=\frac{1}{2}\times B\times H$........(1)

• Here, $A_T$ is the area of the triangle, B is the Base and H is the height of the triangle.

Calculation:

Consider the figure shown below:

  

There are two rectangles and one triangle in the above figure.

The total area of the above figure is sum of the areas of the two triangles and one triangle.

Area of the rectangle with length $12\text{ in}$ and breadth $8\text{ in}$ is,

  $\begin{array}{c}A=L\times B\\ =12\times 8\\ =96{\text{in}}^2\end{array}$

Area of the triangle with length $7\text{ in}$ and base $5\text{ in}$ is,

  $\begin{array}{c}{A}_T=\frac{1}{2}\times B\times H\\ =\frac{1}{2}\times 5\times 7\\ =\frac{35}{2}\\ =17.5{\text{in}}^2\end{array}$

Area of the triangle with height $7\text{ in}$ and breadth $5\text{ in}$ is,

  $\begin{array}{c}A=L\times B\\ =7\times 5\\ =35{\text{in}}^2\end{array}$

Total area of the figure is,

  $\begin{array}{c}A=A_1+A_2+A_T\\ =96{\text{in}}^2+35{\text{in}}^2+17.5{\text{in}}^2\\ =148.5{\text{in}}^2\end{array}$

Hence, the area of the figure is $148.5{\text{ in}}^2$.