Similar Triangles and Proportions - GED Math

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Question

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

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

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Answer

$\Delta CBA \sim \Delta FED$ 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.

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Question

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

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Answer

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.

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Question

Note: Figure NOT drawn to scale.

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

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Answer

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

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Question

Note: Figures NOT drawn to scale.

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

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Answer

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$

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Question

Note: Figures NOT drawn to scale.

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

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Answer

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.

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}$ = $\frac{45}{10}$ = 4.5$ .

This can be used to find and :

$$\frac{DE}{AB}$ = 4.5$

$$\frac{DE}{8}$ = 4.5$

$DE = 8 \cdot 4.5 = 36$

$$\frac{EF}{BC}$ = 4.5$

$$\frac{EF}{6}$ = 4.5$

$EF = 6 \cdot 4.5 = 27$

The area of $\Delta DE F$ is therefore

$$\frac{1}{2}$ \cdot DE \cdot EF = $\frac{1}{2}$ \cdot 36 \cdot 27 =486$ .

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Question

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

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Answer

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 .

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Question

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

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

Tap to reveal answer

Answer

$\Delta CBA \sim \Delta FED$ 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.

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Question

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

Tap to reveal answer

Answer

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.

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Question

Note: Figure NOT drawn to scale.

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

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Answer

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

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Question

Note: Figures NOT drawn to scale.

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

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Answer

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$

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Question

Note: Figures NOT drawn to scale.

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

Tap to reveal answer

Answer

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.

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}$ = $\frac{45}{10}$ = 4.5$ .

This can be used to find and :

$$\frac{DE}{AB}$ = 4.5$

$$\frac{DE}{8}$ = 4.5$

$DE = 8 \cdot 4.5 = 36$

$$\frac{EF}{BC}$ = 4.5$

$$\frac{EF}{6}$ = 4.5$

$EF = 6 \cdot 4.5 = 27$

The area of $\Delta DE F$ is therefore

$$\frac{1}{2}$ \cdot DE \cdot EF = $\frac{1}{2}$ \cdot 36 \cdot 27 =486$ .

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Question

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

Tap to reveal answer

Answer

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 .

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Question

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

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

Tap to reveal answer

Answer

$\Delta CBA \sim \Delta FED$ 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.

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Question

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

Tap to reveal answer

Answer

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.

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Question

Note: Figure NOT drawn to scale.

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

Tap to reveal answer

Answer

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

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Question

Note: Figures NOT drawn to scale.

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

Tap to reveal answer

Answer

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$

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Question

Note: Figures NOT drawn to scale.

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

Tap to reveal answer

Answer

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.

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}$ = $\frac{45}{10}$ = 4.5$ .

This can be used to find and :

$$\frac{DE}{AB}$ = 4.5$

$$\frac{DE}{8}$ = 4.5$

$DE = 8 \cdot 4.5 = 36$

$$\frac{EF}{BC}$ = 4.5$

$$\frac{EF}{6}$ = 4.5$

$EF = 6 \cdot 4.5 = 27$

The area of $\Delta DE F$ is therefore

$$\frac{1}{2}$ \cdot DE \cdot EF = $\frac{1}{2}$ \cdot 36 \cdot 27 =486$ .

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Question

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

Tap to reveal answer

Answer

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 .

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

$\Delta CBA \sim \Delta FED$ 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.

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Question

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

Tap to reveal answer

Answer

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.

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