Chapter 11.1, Problem 6PSA

a

To determine

To find : The area of figure shown in part a

Answer to Problem 6PSA

The area is $A=124$

Explanation of Solution

Given information : The following information has been given

Length of bigger rectangle $l_1=12$

Width of bigger rectangle $w_1=12$

Length of smaller rectangle $l_2=5$

Width of smaller rectangle $w_2=4$

Formula used : The area of any rectangle will be

  $A=(lengt{h}_1)(widt{h}_1)$

Calculation : We know that the area of any rectangle will be

  $A=(lengt{h}_1)(widt{h}_1)-(lengt{h}_2)(widt{h}_2)$

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

  $A=(12)(12)-(5)(4)=124$

Thus, we get

  $A=124$

b

To determine

To find : The area of figure shown in part b

Answer to Problem 6PSA

The area is $A=102$

Explanation of Solution

Given information : The following information has been given

Length of bigger rectangle $l_1=14$

Width of bigger rectangle $w_1=9$

Length of smaller rectangle $l_2=4$

Width of smaller rectangle $w_2=6$

Formula used : The area of any rectangle will be

  $A=(lengt{h}_1)(widt{h}_1)$

Calculation : We know that the area of any rectangle will be

  $A=(lengt{h}_1)(widt{h}_1)-(lengt{h}_2)(widt{h}_2)$

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

  $A=(14)(9)-(4)(6)=102$

Thus, we get

  $A=102$

c

To determine

To find : The area of figure shown in part c

Answer to Problem 6PSA

The area is $A=116$

Explanation of Solution

Given information : The following information has been given

Length of bigger rectangle $l_1=20$

Width of bigger rectangle $w_1=10$

Length of smaller rectangle $l_2=14$

Width of smaller rectangle $w_2=6$

Formula used : The area of any rectangle will be

  $A=(lengt{h}_1)(widt{h}_1)$

Calculation : We know that the area of any rectangle will be

  $A=(lengt{h}_1)(widt{h}_1)-(lengt{h}_2)(widt{h}_2)$

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

  $A=(20)(10)-(14)(6)=116$

Thus, we get

  $A=116$