Chapter 11, Problem 40SGR

To determine

To calculate the area.

Answer to Problem 40SGR

  $\frac{33}{4}i{n}^2$

Explanation of Solution

Given Information: $1\frac{1}{2}in,2in,3in.$ .

Formula Used:Area of triangle = $\frac{1}{2}\times Base\times Height$ & Area of rectangle = $Length\times Breadth$

Calculation:

In given figure, it shows 2 right angled triangle and 1 rectangle.

So, Area of figure = Area of 2 right angled triangle + Area of rectangle.

So, first calculate area of right angled triangle.

Right angled triangle has 3 sides and $90°$ angle. Hence,

Hence, $1\frac{1}{2}in=\frac{3}{2}in$

Area of 1 right angled triangle = $\frac{1}{2}\times Base\times Height$

Area of 1 right angled triangle = $\frac{1}{2}\times \frac{3}{2}\times \frac{3}{2}i{n}^2$

Area of 1 right angled triangle = $\frac{9}{8}i{n}^2$

Area of 2 right angled triangle = $\frac{9}{8}\times 2i{n}^2$

Area of 2 right angled triangle = $\frac{9}{4}i{n}^2$

Now, calculate area of rectangle = $Length\times Breadth$

Length = 3 in and Breadth = 2 in.

Area of rectangle = $3\times 2i{n}^2$

Area of rectangle = $6i{n}^2$

Area of figure = Area of 2 right angled triangle + Area of rectangle

Area of figure = $\frac{9}{4}i{n}^2+6i{n}^2$

Area of figure = $\frac{33}{4}i{n}^2$