How to find if right triangles are similar - Geometry

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $\angle F$ is a right angle; $\overline{AB}=20$ ; $\overline{AC}=10$ ; $\overline{DE}=30$

Find $\overline{EF}$ .

Answer

Since $\bigtriangleup ABC \sim \bigtriangleup DEF$ and $\angle F$ is a right angle, $\angle C$ is also a right angle.

$\overline{AB}$ is the hypotenuse of the first triangle; since one of its legs $\overline{AC}$ is half the length of that hypotenuse, $\bigtriangleup ABC$ is 30-60-90 with $\overline{AC}$ the shorter leg and $\overline{BC}$ the longer.

Because the two are similar triangles, $\overline{DE}$ is the hypotenuse of the second triangle, and $\overline{EF}$ is its longer leg.

Therefore, $EF = \frac{\sqrt{3}}{2} \cdot DE = \frac{\sqrt{3}}{2} \cdot 30 = 15 \sqrt{3}$ .

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Question

Which of the following is sufficient to say that two right triangles are similar?

Answer

If all three angles of a triangle are congruent but the sides are not, then one of the triangles is a scaled up version of the other. When this happens the proportions between the sides still remains unchanged which is the criteria for similarity.

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Question

Which of the following statements is true regarding the two triangles?

Answer

Though we must do a little work, we can show these triangles are similar. First, right triangles are not necessarily always similar. They must meet the necessary criteria like any other triangles; furthermore, there is no Hypotenuse-Leg Theorem for similarity, only for congruence; therefore, we can eliminate two answer choices.

However, we can use the Pythagorean Theorem with the smaller triangle to find the missing leg. Doing so gives us a length of 48. Comparing the ratio of the shorter legs in each trangle to the ratio of the longer legs we get

$\frac{20}{10}=\frac{2}{1}$

$\frac{48}{24}=\frac{2}{1}$

In both cases, the leg of the larger triangle is twice as long as the corresponding leg in the smaller triangle. Given that the angle between the two legs is a right angle in each triangle, these angles are congruent. We now have enough evidence to conclude similarity by Side-Angle-Side.

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Question

Two triangles, $\Delta A$ and $\Delta B$ , are similar when:

Answer

The Similar Figures Theorem holds that similar figures have both equal corresponding angles and proportional corresponding lengths. Either condition alone is not sufficient. If two figures have both equal corresponding angles and equal corresponding lengths then they are congruent, not similar.

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Question

$\Delta ABC$ and $\Delta DEF$ are triangles.

$m\angle B=90^{\circ}$

$m\angle E=90^{\circ}$

Are $\Delta ABC$ and $\Delta DEF$ similar?

Answer

The Similar Figures Theorem holds that similar figures have both equal corresponding angles and proportional corresponding lengths. In other words, we need to know both the measures of the corresponding angles and the lengths of the corresponding sides. In this case, we know only the measures of $\angle B$ and $\angle E$ . We don't know the measures of any of the other angles or the lengths of any of the sides, so we cannot answer the question -- they might be similar, or they might not be.

It's not enough to know that both figures are right triangles, nor can we assume that angles are the same measurement because they appear to be.

Similar triangles do not have to be the same size.

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Question

$\Delta ABC$ and $\Delta DEF$ are similar triangles.

$m\angle A=30^{\circ}, m\angle B=90^{\circ}, m\angle C=60^{\circ}$

$\overline{AB}=\sqrt{3}, \overline{BC}=1, \overline{AC}=2$

$m\angle D=30^{\circ}, m\angle E=90^{\circ}, m\angle F=60^{\circ}$

$\overline{EF}=2$

What is the length of $\overline{DE}$ ?

Answer

Since $\Delta ABC$ and $\Delta DEF$ are similar triangles, we know that they have proportional corresponding lengths. We must determine which sides correspond. Here, we know $\overline{EF}$ corresponds to $\overline {BC}$ because both line segments lie opposite $30^{\circ}$ angles and between $90^{\circ}$ and $60^{\circ}$ angles. Likewise, we know $\overline{DE}$ corresponds to $\overline {AB}$ because both line segments lie opposite $60^{\circ}$ angles and between $30^{\circ}$ and $90^{\circ}$ angles. We can use this information to set up a proportion and solve for the length of $\overline{DE}$ .

$\frac{\overline{EF}}{\overline{BC}}=\frac{\overline{DE}}{\overline{AB}}$

Substitute the known values.

$\frac{2}{1}=\frac{\overline{DE}}{\sqrt{3}}$

Cross-multiply and simplify.

$2\sqrt{3}=\overline{DE}$

and result from setting up an incorrect proportion. $\sqrt{6}$ results from incorrectly multiplying and $\sqrt{3}$ .

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Question

Are these triangles similar? If so, list the scale factor.

Answer

The two triangles are similar, but we can't be sure of that until we can compare all three corresponding pairs of sides and make sure the ratios are the same. In order to do that, we first have to solve for the missing sides using the Pythagorean Theorem.

$\small 9^2 + 12^2 = c^2$

$\small 81 + 144 = c^2$

$\small 225 = c^2$

$\small \sqrt{225 } = 15 = c$

The smaller triangle is missing not the hypotenuse, c, but one of the legs, so we'll use the formula slightly differently.

$\small x^2 + 6^2 = 10^2$

$\small x^2 + 36 = 100$ subtract 36 from both sides

$\small x^2 = 64$

$\small x = \sqrt{64} = 8$

Now we can compare all three ratios of corresponding sides:

$\small \frac{6}{9 } =? \frac{8}{12} =? \frac{10}{15}$ one way of comparing these ratios is to simplify them.

We can simplify the leftmost ratio by dividing top and bottom by 3 and getting $\small \frac{2}{3}$ .

We can simplify the middle ratio by dividing top and bottom by 4 and getting $\small \frac{2}{3}$ .

Finally, we can simplify the ratio on the right by dividing top and bottom by 5 and getting $\small \frac{2}{3}$ .

This means that the triangles are definitely similar, and $\small \frac{2}{3}$ is the scale factor.

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Question

Are these right triangles similar? If so, state the scale factor.

Answer

In order to compare these triangles and determine if they are similar, we need to know all three side lengths in both triangles. To get the missing ones, we can use Pythagorean Theorem:

$\small 15^2 + 8^2 = c^2$

$\small 225 + 64 = c^2$

$\small 289 = c^2$ take the square root

$\small \sqrt{289} = 17 = c$

The other triangle is missing one of the legs rather than the hypotenuse, so we'll adjust accordingly:

$\small 6^2 + x^2 = 10^2$

$\small 36 + x^2 = 100$ subtract 36 from both sides

$\small x^2 = 64$

$\small x= 8$

Now we can compare ratios of corresponding sides:

$\small \frac{8}{6} =? \frac{15}{8 } =? \frac{17}{10}$

The first ratio simplifies to $\small \frac{4}{3}$ , but we can't simplify the others any more than they already are. The three ratios clearly do not match, so these are not similar triangles.

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Question

Are these triangles similar? Give a justification.

Answer

These triangles were purposely drawn misleadingly. Just from glancing at them, the angles that appear to correspond are given different angle measures, so they don't "look" similar. However, if we subtract, we figure out that the missing angle in the triangle with the 66-degree angle must be 24 degrees, since $\small \small 90- 66 =24$ . Similarly, the missing angle in the triangle with the 24-degree angle must be 66 degress. This means that all 3 corresponding pairs of angles are congruent, making the triangles similar.

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Question

Refer to the above figure.

True, false, or inconclusive: $\bigtriangleup ADB \sim \bigtriangleup BDC$ .

Answer

$\overline{AD}$ is an altitude of $\bigtriangleup ABC$ , so it divides the triangle into two smaller triangles similar to each other - that is, if we match the shorter legs, the longer legs, and the hypotenuses, the similarity statement is

$\bigtriangleup ADB \sim \bigtriangleup BDC$ .

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Question

Refer to the above diagram.

True or false: $\bigtriangleup AOB \sim \bigtriangleup COD$

Answer

The distance from the origin to is the absolute value of the -coordinate of , which is . Similarly, , , and . Also, since the axes intersect at right angles, $\angle AOB$ and $\angle COD$ are both right, and, consequently, congruent.

According to the Side-Angle-Side Similarity Theorem (SASS), if two sides of a triangle are in proportion to the corresponding sides of a second triangle, and their included angles are congruent, the triangles are similar.

We can test the proportion statement

$\frac{OA}{OC} = \frac{OB}{OD}$

by substituting:

$\frac{12}{16} = \frac{6}{8}$

Test the truth of this statement by comparing their cross products:

$12 \times 8 = 6 \times 16$

The cross-products are equal, making the proportion statement true, so two pairs of sides are in proportion. Also, their included angles $\angle AOB$ and $\angle COD$ are congruent. This sets up the conditions of SASS, so $\bigtriangleup AOB \sim \bigtriangleup COD$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ .

$\angle B$ and $\angle E$ are both right angles.

$\overline{AB} \cong \overline{BC}$

$\overline{DE} \cong \overline{EF}$

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

Answer

If we seek to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond to , , and , respectively.

By the Side-Angle-Side Similarity Theorem (SASS), if two sides of a triangle are in proportion with the corresponding sides of another triangle, and the included angles are congruent, then the triangles are similar.

and , so by the Division Property of Equality, $\frac{AB}{DE} = \frac{BC}{EF}$ . Also, $\angle B$ and $\angle E$ , their respective included angles, are both right angles, so $\angle B \cong \angle E$ . The conditions of SASS are met, so $\bigtriangleup ABC \sim \bigtriangleup DEF$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $\angle F$ is a right angle; $\overline{AB}=20$ ; $\overline{AC}=10$ ; $\overline{DE}=30$

Find $\overline{EF}$ .

Answer

Since $\bigtriangleup ABC \sim \bigtriangleup DEF$ and $\angle F$ is a right angle, $\angle C$ is also a right angle.

$\overline{AB}$ is the hypotenuse of the first triangle; since one of its legs $\overline{AC}$ is half the length of that hypotenuse, $\bigtriangleup ABC$ is 30-60-90 with $\overline{AC}$ the shorter leg and $\overline{BC}$ the longer.

Because the two are similar triangles, $\overline{DE}$ is the hypotenuse of the second triangle, and $\overline{EF}$ is its longer leg.

Therefore, $EF = \frac{\sqrt{3}}{2} \cdot DE = \frac{\sqrt{3}}{2} \cdot 30 = 15 \sqrt{3}$ .

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Question

Which of the following is sufficient to say that two right triangles are similar?

Answer

If all three angles of a triangle are congruent but the sides are not, then one of the triangles is a scaled up version of the other. When this happens the proportions between the sides still remains unchanged which is the criteria for similarity.

Compare your answer with the correct one above

Question

Which of the following statements is true regarding the two triangles?

Answer

Though we must do a little work, we can show these triangles are similar. First, right triangles are not necessarily always similar. They must meet the necessary criteria like any other triangles; furthermore, there is no Hypotenuse-Leg Theorem for similarity, only for congruence; therefore, we can eliminate two answer choices.

However, we can use the Pythagorean Theorem with the smaller triangle to find the missing leg. Doing so gives us a length of 48. Comparing the ratio of the shorter legs in each trangle to the ratio of the longer legs we get

$\frac{20}{10}=\frac{2}{1}$

$\frac{48}{24}=\frac{2}{1}$

In both cases, the leg of the larger triangle is twice as long as the corresponding leg in the smaller triangle. Given that the angle between the two legs is a right angle in each triangle, these angles are congruent. We now have enough evidence to conclude similarity by Side-Angle-Side.

Compare your answer with the correct one above

Question

Two triangles, $\Delta A$ and $\Delta B$ , are similar when:

Answer

The Similar Figures Theorem holds that similar figures have both equal corresponding angles and proportional corresponding lengths. Either condition alone is not sufficient. If two figures have both equal corresponding angles and equal corresponding lengths then they are congruent, not similar.

Compare your answer with the correct one above

Question

$\Delta ABC$ and $\Delta DEF$ are triangles.

$m\angle B=90^{\circ}$

$m\angle E=90^{\circ}$

Are $\Delta ABC$ and $\Delta DEF$ similar?

Answer

The Similar Figures Theorem holds that similar figures have both equal corresponding angles and proportional corresponding lengths. In other words, we need to know both the measures of the corresponding angles and the lengths of the corresponding sides. In this case, we know only the measures of $\angle B$ and $\angle E$ . We don't know the measures of any of the other angles or the lengths of any of the sides, so we cannot answer the question -- they might be similar, or they might not be.

It's not enough to know that both figures are right triangles, nor can we assume that angles are the same measurement because they appear to be.

Similar triangles do not have to be the same size.

Compare your answer with the correct one above

Question

$\Delta ABC$ and $\Delta DEF$ are similar triangles.

$m\angle A=30^{\circ}, m\angle B=90^{\circ}, m\angle C=60^{\circ}$

$\overline{AB}=\sqrt{3}, \overline{BC}=1, \overline{AC}=2$

$m\angle D=30^{\circ}, m\angle E=90^{\circ}, m\angle F=60^{\circ}$

$\overline{EF}=2$

What is the length of $\overline{DE}$ ?

Answer

Since $\Delta ABC$ and $\Delta DEF$ are similar triangles, we know that they have proportional corresponding lengths. We must determine which sides correspond. Here, we know $\overline{EF}$ corresponds to $\overline {BC}$ because both line segments lie opposite $30^{\circ}$ angles and between $90^{\circ}$ and $60^{\circ}$ angles. Likewise, we know $\overline{DE}$ corresponds to $\overline {AB}$ because both line segments lie opposite $60^{\circ}$ angles and between $30^{\circ}$ and $90^{\circ}$ angles. We can use this information to set up a proportion and solve for the length of $\overline{DE}$ .

$\frac{\overline{EF}}{\overline{BC}}=\frac{\overline{DE}}{\overline{AB}}$

Substitute the known values.

$\frac{2}{1}=\frac{\overline{DE}}{\sqrt{3}}$

Cross-multiply and simplify.

$2\sqrt{3}=\overline{DE}$

and result from setting up an incorrect proportion. $\sqrt{6}$ results from incorrectly multiplying and $\sqrt{3}$ .

Compare your answer with the correct one above

Question

Are these triangles similar? If so, list the scale factor.

Answer

The two triangles are similar, but we can't be sure of that until we can compare all three corresponding pairs of sides and make sure the ratios are the same. In order to do that, we first have to solve for the missing sides using the Pythagorean Theorem.

$\small 9^2 + 12^2 = c^2$

$\small 81 + 144 = c^2$

$\small 225 = c^2$

$\small \sqrt{225 } = 15 = c$

The smaller triangle is missing not the hypotenuse, c, but one of the legs, so we'll use the formula slightly differently.

$\small x^2 + 6^2 = 10^2$

$\small x^2 + 36 = 100$ subtract 36 from both sides

$\small x^2 = 64$

$\small x = \sqrt{64} = 8$

Now we can compare all three ratios of corresponding sides:

$\small \frac{6}{9 } =? \frac{8}{12} =? \frac{10}{15}$ one way of comparing these ratios is to simplify them.

We can simplify the leftmost ratio by dividing top and bottom by 3 and getting $\small \frac{2}{3}$ .

We can simplify the middle ratio by dividing top and bottom by 4 and getting $\small \frac{2}{3}$ .

Finally, we can simplify the ratio on the right by dividing top and bottom by 5 and getting $\small \frac{2}{3}$ .

This means that the triangles are definitely similar, and $\small \frac{2}{3}$ is the scale factor.

Compare your answer with the correct one above

Question

Are these right triangles similar? If so, state the scale factor.

Answer

In order to compare these triangles and determine if they are similar, we need to know all three side lengths in both triangles. To get the missing ones, we can use Pythagorean Theorem:

$\small 15^2 + 8^2 = c^2$

$\small 225 + 64 = c^2$

$\small 289 = c^2$ take the square root

$\small \sqrt{289} = 17 = c$

The other triangle is missing one of the legs rather than the hypotenuse, so we'll adjust accordingly:

$\small 6^2 + x^2 = 10^2$

$\small 36 + x^2 = 100$ subtract 36 from both sides

$\small x^2 = 64$

$\small x= 8$

Now we can compare ratios of corresponding sides:

$\small \frac{8}{6} =? \frac{15}{8 } =? \frac{17}{10}$

The first ratio simplifies to $\small \frac{4}{3}$ , but we can't simplify the others any more than they already are. The three ratios clearly do not match, so these are not similar triangles.

Compare your answer with the correct one above

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