Chapter 7.4, Problem 11CE

(a)

To determine

To Complete: the statement $\Delta IKJ\sim \underset{\_}{?}$ for the figure shown.

Answer to Problem 11CE

  $\Delta HLJ$

Explanation of Solution

Given:

Figure is shown below.

  

Calculation:

If two triangles are similar, then ratio of their corresponding sides are equal.

  $\Delta IKJ$ could be similar to $\Delta HLJ$ for the fixed value of when equating with scale factors.

Therefore , the answer is $\Delta HLJ$ .

(c)

To determine

To Complete: the statement $\frac{?}{x}=\frac{9}{24}\text{ and }x=?$ for the figure shown.

Answer to Problem 11CE

  $12$ and $32$

Explanation of Solution

Given:

Figure is shown below.

  

Calculation:

Since, $\Delta IKJ\sim \Delta HLJ$

Here, ratio of corresponding sides are.

  $\begin{array}{l}\frac{IK}{HL}=\frac{JK}{JL}\\ \frac{12}{x}=\frac{9}{24}\\ x=\frac{12\times 24}{9}\\ x=32\end{array}$ .

Therefore, the answer is $12$ and $32$

(c)

To determine

To Complete: the statement $\frac{9}{24}=\frac{6}{?}\text{ and y=?}$ for the figure shown.

Answer to Problem 11CE

  $16$ and $10$

Explanation of Solution

Given:

Figure is shown below.

  

  $\frac{9}{24}=\frac{6}{?}\text{ and y=?}$

Calculation:

Since, $\Delta IKJ\sim \Delta HLJ$

Here, ratio of corresponding sides are.

  $\begin{array}{l}\frac{JK}{JL}=\frac{JI}{JH}\\ \frac{9}{24}=\frac{6}{6+y}\\ \frac{3}{8}=\frac{6}{6+y}\\ \frac{1}{8}=\frac{2}{6+y}\\ \left(6+y\right)=2\times 8\\ 6+y=16\\ y=10\end{array}$

And

  $\begin{array}{l}\frac{9}{24}=\frac{6}{6+y}\\ \frac{9}{24}=\frac{6}{6+10}\\ \frac{9}{24}=\frac{6}{16}\end{array}$

Therefore, the answer is $16$ and $10$