Question

exercises 9-12
in exercises 9-12, use the similar triangles abc and abc shown to determine the following.

1. the length of bc
2. the length of ab
3. m∠bac
4. m∠abc

Explanation:

Step1: Recall property of similar - triangles

For similar triangles $\triangle ABC$ and $\triangle A'B'C'$, the ratios of corresponding sides are equal and corresponding angles are congruent.

Step2: Find the length of $BC$

The ratio of corresponding sides of similar triangles is $\frac{AA' + A'C}{A'C}=\frac{8 + 4}{4}=3$. Since $\frac{BC}{B'C'}=3$ and $B'C' = 3.4$ in, then $BC=3\times3.4 = 10.2$ in.

Step3: Find the length of $A'B'$

We know that $\frac{AB}{A'B'}=\frac{AC}{A'C}$. Given $AB = 6$ in, and the ratio of side - lengths $\frac{AC}{A'C}=3$. Let $A'B'=x$, then $\frac{6}{x}=3$, so $x = 2$ in.

Step4: Find $m\angle BAC$

Since corresponding angles of similar triangles are congruent, and in $\triangle A'B'C'$, $m\angle B'A'C'=30^{\circ}$, then $m\angle BAC = 30^{\circ}$.

Step5: Find $m\angle ABC$

First, find the third angle in $\triangle A'B'C'$. Using the angle - sum property of a triangle ($m\angle A'+m\angle B'+m\angle C'=180^{\circ}$), we have $m\angle A' = 30^{\circ}$, $m\angle B'=88^{\circ}$, so $m\angle C'=180-(30 + 88)=62^{\circ}$. Since corresponding angles of similar triangles are congruent, $m\angle ABC=m\angle A'B'C'=88^{\circ}$.

Answer:

1. $10.2$ in
2. $2$ in
3. $30^{\circ}$
4. $88^{\circ}$