Chapter 11.3, Problem 7PSB

a

To determine

To find : The area of figure shown in part a

Answer to Problem 7PSB

The area is $A=396$

Explanation of Solution

Given information : The following information has been given

Length of rectangle $l_1=22$

Width of rectangle $w_1=9$

Parallel sides of trapezoid are $a=22$ and $b=14$

Height of trapezoid $h=11$

Formula used : The area of any trapezoid will be

  $A=\frac{a+b}{2}h$

Calculation : We know that the area of the given combination will be

  $A=(l_1)\times (w_1)+\frac{a+b}{2}h$

Hence, substituting the values from the given information, we get

  $A=(22)\times (9)+\frac{1}{2}(22+14)\times (11)=396$

Thus, we get

  $A=396$

b

To determine

To find : The area of figure shown in part b

Answer to Problem 7PSB

The area is $A=79.5$

Explanation of Solution

Given information : The following information has been given

Parallel sides of trapezoid are $a=9$ and $b=12$

Height of trapezoid $h=3$

Altitude of triangle $l=\sqrt{{10}^2-6^2}=8$

Base of triangle $w=12$

Formula used : The area of any trapezoid will be

  $A=\frac{a+b}{2}h$

Calculation : We know that the area of the given combination will be

  $A=\frac{1}{2}(l)\times (w)+\frac{a+b}{2}h$

Hence, substituting the values from the given information, we get

  $A=\frac{1}{2}(8)\times (12)+\frac{1}{2}(21)\times (3)=79.5$

Thus, we get

  $A=79.5$