Chapter 11.6, Problem 12PPS

To determine

To Find: The area of the given figure.

Answer to Problem 12PPS

  $A=399\text{ sq}\text{. cm}$

Explanation of Solution

Given information:

The given figure is

Concept and Formula Used:

Area of triangle $A=\frac{1}{2}\left(\text{base}\times \text{height}\right)$ & Area of trapezoid $A=\frac{1}{2}h\left(a+b\right)$

Calculation:

The given figure is combination of triangle and trapezoid.

To find the area of the figure add the area of triangle and trapezoid.

The sides of the triangle is $base=\left(28.5-16.9\right)cm=11.6cm\&height=14cm$ and the bases of trapezoid is equal to $a=28.5cm\&b=16.9cm$ and height of trapezoid is $h=14cm$

Area of figure is sum of area of rectangle and trapezoid.

  $\begin{array}{l}\text{Area}=\text{Area of triangle}+\text{Area of trapezoid}\\ A=\frac{1}{2}\left(\text{base}\times \text{height}\right)+\frac{1}{2}h\left(a+b\right)\\ \text{where }base=11.6cm,height=14cm,a=28.5cm,b=16.9cm\&h=14cm\\ A=\frac{1}{2}\left(11.6\times 14\right)+\frac{1}{2}\left(14\right)\left(28.5+16.9\right)\\ A=81.2+\frac{1}{2}\left(14\right)\left(45.4\right)\\ A=81.2+317.8\\ A=399\text{ sq}\text{. cm}\end{array}$

Conclusion:

  $A=399\text{ sq}\text{. cm}$