Chapter 11, Problem 32RP

a.

To determine

To calculate: The area ratio of region ⏸ and region ⏸.

Answer to Problem 32RP

The area ratio is $5:4$ .

Explanation of Solution

Given information:

Side AC = 9,

Side BC = 6.

Formula used:

The below similar figures theorem is used:

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of ratio of their corresponding sides. This proves that the ratio of the area of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

Calculation:

  

Triangle in region ⏸ is similar to triangle in region ⏸. These two triangles are similar to triangle in region ⏸ and ⏸.

  $\begin{array}{l}\frac{A_{\mathrm{Ⅰ}}}{A_{\mathrm{Ⅱ}}}={\left(\frac{s_1}{s_2}\right)}^2\\ \frac{A_{\mathrm{Ⅱ}}}{A_{\mathrm{Ⅰ}}+A_{\mathrm{Ⅱ}}}={\left(\frac{BC}{AC}\right)}^2={\left(\frac{6}{9}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}----Equation1\\ \frac{A_{\mathrm{Ⅰ}}}{A_{\mathrm{Ⅰ}}+A_{\mathrm{Ⅱ}}}=\frac{9}{9}-\frac{4}{9}=\frac{5}{9}----Equation2\end{array}$

Dividing Equation 1 by Equation 2, we get

  $\frac{A_{\mathrm{Ⅰ}}}{A_{\mathrm{Ⅱ}}}=\left(\frac{\frac{5}{9}}{\frac{4}{9}}\right)=\frac{5}{4}$

b.

To determine

To find: The area ratio of region ⏸ and region ⏸.

Answer to Problem 32RP

The area ratio is $25:39$ .

Explanation of Solution

Given information:

Side QS = 10,

Side SP = 6.

Formula used:

The below similar figures theorem is used:

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of ratio of their corresponding sides. This proves that the ratio of the area of two similar triangles is proportional to the squares of the corresponding sides of both the triangles.

Calculation:

  

Triangle in region ⏸ is similar to triangle in region ⏸. These two triangles are similar to triangle in region ⏸ and ⏸.

  $\begin{array}{l}\frac{A_{\mathrm{Ⅰ}}}{A_{\mathrm{Ⅰ}}+A_{\mathrm{Ⅱ}}}={\left(\frac{QS}{QP}\right)}^2={\left(\frac{5}{8}\right)}^2=\frac{25}{64}----Equation1\\ \frac{A_{\mathrm{Ⅱ}}}{A_{\mathrm{Ⅰ}}+A_{\mathrm{Ⅱ}}}=\frac{64}{64}-\frac{25}{64}=\frac{39}{64}----Equation2\end{array}$

Dividing Equation 1 by Equation 2, we get

  $\frac{A_{\mathrm{Ⅰ}}}{A_{\mathrm{Ⅱ}}}=\left(\frac{\frac{25}{64}}{\frac{39}{64}}\right)=\frac{25}{39}=25:39$