GED Math - Similar Triangles and Proportions

In the figure below, the two triangles are similar. Find the value of .

Explanation

Since the two triangles are similar, we know that their corresponding sides must be in the same ratio to each other. Thus, we can write the following equation:

$\frac{12}{8}=\frac{9}{x}$

Now, solve for .

Triangles

Note: Figure NOT drawn to scale.

Refer to the above diagram. If $\Delta ABC \sim \Delta DE F$ , which of the following is false?

$\angle E$ is a right angle

$\angle A \cong \angle D$

Explanation

Suppose $\Delta ABC \sim \Delta DE F$ .

Corresponding angles of similar triangles are congruent, so $\angle A \cong \angle D$ . Also, $\angle B \cong \angle E$ , so, since $\angle B$ is a right angle, so is $\angle E$ .

Corresponding sides of similar triangles are in proportion. Since

$\frac{DE}{AB} = \frac{24}{8} = 3$ ,

the similarity ratio of $\Delta DE F$ to $\Delta ABC$ is 3.

By the Pythagorean Theorem, since is the hypotenuse of a right triangle with legs 6 and 8, its measure is

$AC = \sqrt{6^{2} + 8 ^{2}} = \sqrt{36 + 64 } = \sqrt{100} = 10$ .

$DF = 3 (AC) = 3 \cdot 10 = 30$ , so is a true statement.

But $EF = 3 (BC) = 3 \cdot 8 = 24 eq 20$ , so is false if the triangles are similar. This is the correct choice.

Which of the following statements is not a consequence of the statement

$\Delta ABC \sim \Delta DE F$ ?

$\angle A \cong \angle D$

$\frac{AB}{DE} = \frac{AC}{DF}$

$\Delta CBA \sim \Delta FED$

Explanation

$\Delta CBA \sim \Delta FED$ is simply a restatement of $\Delta ABC \sim \Delta DE F$ , since the names of the corresponding vertices of the similar triangles are still in the same relative positions.

$\angle A \cong \angle D$ is a consequence of $\Delta ABC \sim \Delta DE F$ , since corresponding angles of similar triangles are, by definition, congruent.

$\frac{AB}{DE} = \frac{AC}{DF}$ is a consequence of $\Delta ABC \sim \Delta DE F$ , since corresponding sides of similar triangles are, by definition, in proportion.

However, similar triangles need not have congruent corresponding sides. Therefore, it does not necessarily follow that . This is the correct choice.

Which of the following statements follows from the statement $\Delta KLM \sim \Delta XYZ$ ?

$\frac{KL}{LM } = \frac{XY}{YZ}$

$\angle K \cong \angle Y$

$\angle L \cong \angle M$

$\overline{KM} \cong \overline{XZ}$

Explanation

The similarity of two triangles implies nothing about the relationship of two angles of the same triangle. Therefore, $\angle L \cong \angle M$ can be eliminated.

The similarity of two triangles implies that corresponding angles between the triangles are congruent. However, because of the positions of the letters, $\angle K$ in $\Delta KLM$ corresponds to $\angle X$ , not $\angle Y$ , in $\Delta XYZ$ , so $\angle K \cong \angle X$ . The statement $\angle K \cong \angle Y$ can be eliminated.

Similarity of two triangles does not imply any congruence between sides of the triangles, so $\overline{KM} \cong \overline{XZ}$ can be eliminated.

Similarity of triangles implies that corresponding sides are in proportion. $\overline{KL}$ and $\overline{LM}$ in $\Delta KLM$ correspond, respectively, to $\overline{XY}$ and $\overline{YZ}$ in $\Delta XYZ$ . Therefore, it follows that $\frac{KL}{LM } = \frac{XY}{YZ}$ , and this statement is the correct choice.

Triangles

Note: Figures NOT drawn to scale.

Refer to the above figures. Given that $\Delta ABC \sim \Delta DE F$ , evaluate .

$52 \frac{1}{2}$

$93\frac{1}{3}$

Explanation

By the Pythagorean Theorem, since is the hypotenuse of a right triangle with legs 6 and 8, its measure is

$AC = \sqrt{6^{2} + 8 ^{2}} = \sqrt{36 + 64 } = \sqrt{100} = 10$ .

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{DF}{AC} = \frac{70}{10} = 7$ .

Likewise,

$\frac{EF}{BC} = 7$

$\frac{EF}{6} = 7$

Triangles

Note: Figures NOT drawn to scale.

Refer to the above figures. Given that $\Delta ABC \sim \Delta DE F$ , give the area of $\Delta DE F$ .

Explanation

Corresponding angles of similar triangles are congruent, so, since $\angle B$ is right, so is $\angle E$ . This makes $\overline{DE}$ and $\overline{EF}$ the legs of a right triangle, so its area is half their product.