Chapter 8.1, Problem 70PFA

a.

To determine

To write an expression for the perimeter of rectangle on the left.

Answer to Problem 70PFA

  $P=8x+8$

Explanation of Solution

Given:

  

Formula used:

Perimeter of rectangle $=2(l+b)$

where l is the length of the rectangle and b is the breadth of the rectangle.

Calculation:

The length of the rectangle in the left is $3x+4.$

The breadth of the rectangle in the left is $x.$

  $\Rightarrow b=x$

Substituting the length and breadth of the rectangle in the formula of the perimeter,

  $\begin{array}{l}\Rightarrow P=2(l+b)\\ \Rightarrow P=2((3x+4)+x)\\ \Rightarrow P=2(3x+4+x)\\ \Rightarrow P=2(4x+4)\\ \Rightarrow P=8x+8\end{array}$

Conclusion:

Therefore, the perimeter of the rectangle is $8x+8$ .

b.

To determine

To write an expression for the perimeter of rectangle on the right.

Answer to Problem 70PFA

  $P=4x^2+8x+12$

Explanation of Solution

Given:

  

Formula used:

Perimeter of rectangle $=2(l+b)$

where l is the length of the rectangle and b is the breadth of the rectangle.

Calculation:

The length of the rectangle in the right is $2x^2+3x+6.$

  $\Rightarrow l=2x^2+3x+6$

The breadth of the rectangle in the right is $x.$

  $\Rightarrow b=x$

Substituting the length and breadth of the rectangle in the formula of the perimeter,

  $\begin{array}{l}\Rightarrow P=2(l+b)\\ \Rightarrow P=2((2x^2+3x+6)+x)\\ \Rightarrow P=2(2x^2+3x+6+x)\\ \Rightarrow P=2(2x^2+4x+6)\\ \Rightarrow P=4x^2+8x+12\end{array}$

Conclusion:

Therefore, the perimeter of the rectangle is $4x^2+8x+12$ .

c.

To determine

To write an expression for the total perimeter of both the triangles.

Answer to Problem 70PFA

  $P=4x^2+16x+20$

Explanation of Solution

Given:

From subpart (a),

  $P=8x+8$

From subpart (b),

  $P=4x^2+8x+12$

Concept used:

To find the total perimeter of two rectangles, add both the perimeters.

Calculation:

The total perimeter of both the rectangles is,

  $\begin{array}{l}P=(8x+8)+(4x^2+8x+12)\\ P=8x+8+4x^2+8x+12\end{array}$

Combining the like terms,

  $\begin{array}{l}P=4x^2+(8x+8x)+(12+8)\\ P=4x^2+16x+20\end{array}$

Conclusion:

The total perimeter of both the rectangles is $4x^2+16x+20.$

d.

To determine

To find the perimeter of the rectangle by joining length of both the rectangles whose width is x .

Answer to Problem 70PFA

  $4x^2+14x+20$

Explanation of Solution

Given:

  

Formula Used:

Perimeter of the rectangle $=2(l+b)$

where l is the length of the rectangle and b is the breadth of the rectangle.

Calculation:

The total length of the rectangle is the total of the length of both the rectangles.

  $\begin{array}{l}\Rightarrow (3x+4)+(2x^2+3x+6)\\ \Rightarrow 3x+4+2x^2+3x+6\end{array}$

Combining the like terms,

  $\begin{array}{l}\Rightarrow 2x^2+(3x+3x)+(6+4)\\ \Rightarrow 2x^2+6x+10\end{array}$

The length of new rectangle is $2x^2+6x+10$

Width of the rectangle is x .

  

Therefore, the perimeter of the rectangle is,

  $\begin{array}{c}P=2(l+b)\\ =2((2x^2+6x+10)+x)\\ =2(2x^2+6x+10+x)\\ =2(2x^2+7x+10)\\ =4x^2+14x+20\end{array}$

Conclusion:

The perimeter of both the rectangle is $4x^2+14x+20.$

e.

To determine

To check if the perimeter of both the rectangles same as the perimeter of the combined rectangles.

Answer to Problem 70PFA

The perimeters are different.

Explanation of Solution

Given:

From subpart (c),

  $P=4x^2+16x+20.$

From subpart (d),

  $P=4x^2+14x+20$

Comparing the perimeters in both subpart (c) and subpart (d),

The perimeters are different.

Conclusion:

Therefore, the perimeters are different.

f.

To determine

To calculator of the perimeter of the new rectangle if $x=1.5\text{ cm}$

Answer to Problem 70PFA

  $P=50\text{cm}$

Explanation of Solution

Given:

From subpart (d),

  $P=4x^2+14x+20$

  $x=1.5$

Calculation:

Substituting $x=1.5$ in $P=4x^2+14x+20$ ,

  $\Rightarrow P=4{(1.5)}^2+14(1.5)+20$

  $\begin{array}{l}\Rightarrow P=4(2.25)+14(1.5)+20\\ \Rightarrow P=9+21+20\\ \Rightarrow P=50\end{array}$

Conclusion:

Therefore, the perimeter of the new triangle is 50 cm if x = 1.5 cm.