Geometry - How to find if right triangles are similar

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

Find $\overline{EF}$ .

$15 \sqrt{3}$

$10 \sqrt{3}$

$30\sqrt{3}$

Explanation

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

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

Their corresponding angles are equal AND their corresponding lengths are proportional.

Their corresponding angles are equal.

Their corresponding lengths are proportional.

Their corresponding angles are equal AND their corresponding lengths are equal.

Explanation

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.

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

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

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

Triangles

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

There is not enough information given to answer this question.

Yes, because $\Delta ABC$ and $\Delta DEF$ are both right triangles.

Yes, because $\Delta ABC$ and $\Delta DEF$ look similar.

No, because $\Delta ABC$ and $\Delta DEF$ are not the same size.

Explanation

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.

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

Sim right tri 1

Yes - scale factor $\small \frac{2}{3}$

Cannot be determined - we need to know all three sides of both triangles

Yes - scale factor $\small \small \frac{5}{6}{}$

Yes-scale factor

Explanation

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.

Triangles

Refer to the above figure.

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

True

False

Inconclusive

Explanation

$\overline{AD}$ 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

$\bigtriangleup ADB \sim \bigtriangleup BDC$ .

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

All the angles are congruent.

Two of the sides are the same.

Two angles and one side are congruent.

Two sides and one angle are congruent.

Explanation

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.

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 $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

True

False

Explanation

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 $\bigtriangleup ABC \sim \bigtriangleup DEF$

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

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

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

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

$\overline{EF}=2$

Triangles_2

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

$2\sqrt{3}$

$\sqrt{6}$

Explanation

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 $\sqrt{3}$ .

Triangles 3

Refer to the above diagram.

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

True

False

Explanation

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 $\bigtriangleup AOB \sim \bigtriangleup COD$ .

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

Sim right tri 2

No - the side lengths are not proportional

Yes - scale factor $\small \frac{2}{3}$

Yes - scale factor $\small \frac{3}{4}$

Not enough information to be determined

Yes - scale factor

Explanation

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.

How to find if right triangles are similar

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation