Chapter 0.8, Problem 22E

To determine

To calculate: The area of the figure.

Answer to Problem 22E

The area of the figure is $\text{23}{\text{.1 cm}}^2$ .

Explanation of Solution

Formula used:

Area $\left(A\right)$ of rectangle is product of its length $\left(l\right)$ and breadth $\left(b\right)$ . It is mathematically denoted by $A=l\times b$ .

Area $\left(A\right)$ of triangle is half into product of its height and base. Let the base be $\left(b\right)$ and the height be $\left(h\right)$ then it is mathematically denoted by $A=\frac{1}{2}\cdot b\cdot h$ .

Calculation:

The figure has a rectangle with length as 5.2 cm and width as 3.5 cm with triangle on one width with base as 2.8 cm and height as 3.5 cm.

Area of the given figure is the sum of area of triangle and area of rectangle.

Recall that area $\left(A\right)$ of triangle is half into product of its height and base. Let the base be $\left(b\right)$ and the height be $\left(h\right)$ then it is mathematically denoted by $A=\frac{1}{2}\cdot b\cdot h$ .

Apply it,

  $\begin{array}{c}A=\frac{1}{2}\cdot b\cdot h\\ =\frac{1}{2}\cdot 2.8\cdot 3.5\\ =4.9\end{array}$

Now, recall that the area $\left(A\right)$ of rectangle is product of its length $\left(l\right)$ and breadth $\left(b\right)$ . It is mathematically denoted by $A=l\times b$ .

  $\begin{array}{c}A=l\times b\\ =5.2\times 3.5\\ =18.2\end{array}$

Now, area of the given figure is calculated as,

  $\begin{array}{c}A=4.9+18.2\\ =23.1\end{array}$

Rounding off to the nearest tenth, we get,

  $A=23.1{\text{cm}}^{\text{2}}$

Thus, the area of the figure is $\text{23}{\text{.1 cm}}^2$ .