Chapter 11.2, Problem 3PSA

a

To determine

To find : The area of figure shown in part a

Answer to Problem 3PSA

The area is $A=35$

Explanation of Solution

Given information : The following information has been given

Length of rectangle $l_1=7$

Width of rectangle $w_1=3$

Length of triangle $l_2=7$

Width of triangle $w_2=4$

Formula used : The area of any triangle will be

  $A=\frac{1}{2}(base)(height)$

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

  $A=(lengt{h}_1)(widt{h}_1)+\frac{1}{2}(lengt{h}_2)(widt{h}_2)$

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

  $A=(7)(3)+\frac{1}{2}(7)(4)=35$

Thus, we get

  $A=35$

b

To determine

To find : The area of figure shown in part b

Answer to Problem 3PSA

The area is $A=114$

Explanation of Solution

Given information : The following information has been given

Length of rectangle $l_1=\sqrt{9^2+{12}^2}=15$

Width of rectangle $w_1=6$

Base of triangle $b=9$

Height of triangle $h=12$

Formula used : The area of any triangle will be

  $A=\frac{1}{2}(base)(height)$

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

  $A=(length)(width)+\frac{1}{2}(base)(height)$

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

  $A=(15)(6)+\frac{1}{2}(9)(12)=114$

Thus, we get

  $A=114$