Chapter 0.8, Problem 21E

To determine

To calculate: The area of the figure

Answer to Problem 21E

The area of the figure is $\text{22}{\text{.1 cm}}^2$ .

Explanation of Solution

Formula used:

Area $\left(A\right)$ of square is square of its side. Let side of the square be of length $\left(l\right)$ . It is mathematically denoted by $A=l^2$ .

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 square with side length 4.1 cm mounted on a triangle with base 4.1 cm and height 2.6 cm.

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

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,

  $\begin{array}{c}A=\frac{1}{2}\cdot b\cdot h\\ =\frac{1}{2}\cdot 4.1\cdot 2.6\\ =5.33\end{array}$

Now, recall that the area $\left(A\right)$ of square is square of its side. Let side of the square be of length $\left(l\right)$ . It is mathematically denoted by $A=l^2$ .

  $\begin{array}{c}A=l^2\\ ={\left(4.1\right)}^2\\ =16.81\end{array}$

Now, area of the given figure is calculated as,

  $\begin{array}{c}A=16.81+5.33\\ =22.14\end{array}$

Rounding off to the nearest tenth, we get,

  $A=22.1{\text{cm}}^{\text{2}}$

Thus, the area of the figure is $\text{22}{\text{.1 cm}}^2$ .