Chapter 8, Problem 34MCQ

a.

To determine

To write the expression for the area of the largest disc.

Answer to Problem 34MCQ

  $\pi \left(x^2+6.5x+10.5625\right)$ and $3.14\left(x^2+6.5x+10.5625\right)$ .

Explanation of Solution

Given:

Radius of the largest disc $=\left(x+3.25\right)$ inches

Formula used:

Square Formula:

  ${\left(a\pm b\right)}^2=a^2\pm 2ab+b^2$

Area of disc $=\pi r^2$

Calculation:

Given, radius of the largest disc $=\left(x+3.25\right)$ inches.

Now,

Area of the largest disc $=\pi r^2$

  $\begin{array}{l}=\pi {\left(x+3.25\right)}^2\\ =\pi \left(x^2+2\times \left(x\right)\times \left(3.25\right)+{\left(3.25\right)}^2\right)\\ =\pi \left(x^2+6.5x+10.5625\right)\end{array}$  [ $\because {\left(a\pm b\right)}^2=a^2\pm 2ab+b^2$ ]

It can also be written as:

Area of the largest disc $=\pi \left(x^2+6.5x+10.5625\right)$

  $=3.14\left(x^2+6.5x+10.5625\right)$

Conclusion:

Therefore, area of the largest disc in two different expressions is $\pi \left(x^2+6.5x+10.5625\right)$ and $3.14\left(x^2+6.5x+10.5625\right)$ .

b.

To determine

To find the area of smallest and largest disc if $x=10.5$ .

Answer to Problem 34MCQ

Area of the largest disc is $189.0625\pi$ sq. inches and area of the smallest disc is $110.25$ sq. inches.

Explanation of Solution

Given:

Radius of the largest disc $=\left(x+3.25\right)$ inches.

Radius of the smallest disc $=x$ inches.

  $x=10.5$

Formula used:

Square Formula:

  ${\left(a\pm b\right)}^2=a^2\pm 2ab+b^2$

Area of disc $=\pi r^2$

Calculation:

Given radius of the largest disc $=\left(x+3.25\right)$ inches and $x=10.5$ .

  $\Rightarrow$ Radius of the largest disc $=10.5+3.25=13.75$ inches.

Now,

Area of the largest disc $=\pi r^2$

  $\begin{array}{l}=\pi {\left(13.75\right)}^2\\ =\pi \left(189.0625\right)\\ =189.0625\pi \end{array}$

Given radius of the smallest disc $=x$ inches and $x=10.5$ .

  $\Rightarrow$ Radius of the smallest disc $=10.5$ inches.

Now,

Area of the smallest disc $=\pi r^2$

  $\begin{array}{l}=\pi {\left(10.5\right)}^2\\ =\pi \left(110.25\right)\\ =110.25\pi \end{array}$

Conclusion:

Therefore, area of the largest disc is $189.0625\pi$ sq. inches and area of the smallest disc is $110.25$ sq. inches.