Math 2 Quiz: Similarity In Right Triangles

11 questions · exam conditions

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Similarity In Right TrianglesQuestion 1 of 11

In right triangle XYZ with right angle at Y, leg YX has length 15 and leg YZ has length 20. The altitude from Y to hypotenuse XZ divides the hypotenuse into two segments. What is the ratio of the longer segment to the shorter segment?

16:9

4:3

5:3

25:16

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Math 2 Quiz

Math 2 Quiz: Similarity In Right Triangles

Practice Similarity In Right Triangles in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Similarity In Right Triangles, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$16:9$ (correct answer)
$4:3$
$5:3$
$25:16$

Explanation: First find the hypotenuse: XZ = √(15² + 20²) = 25. The segments are found using similarity: if the segments are p and q, then p = 20²/25 = 16 and q = 15²/25 = 9. The ratio of longer to shorter is 16:9. Choice B is the ratio of the legs. Choice C incorrectly uses 20:15 simplified. Choice D uses 25:16 from hypotenuse to longer segment.

Question 2

Triangle RST has a right angle at S. When an altitude is drawn from S to hypotenuse RT, it creates two segments on RT with lengths in the ratio 4:9. If the shorter segment has length 12, what is the length of the altitude from S to RT?

$15$
$24$
$20$
$18$ (correct answer)

Explanation: When you encounter a right triangle with an altitude drawn to the hypotenuse, you're dealing with geometric mean relationships. This altitude creates two smaller right triangles that are similar to each other and to the original triangle. First, find the lengths of both segments. Since they're in the ratio 4:9 and the shorter segment is 12, the longer segment must be

12 \times \frac{9}{4} = 27

. So the two segments have lengths 12 and 27. The key relationship here is that the altitude to the hypotenuse equals the geometric mean of the two segments it creates. This gives us:

h = \sqrt{12 \times 27} = \sqrt{324} = 18

. You can verify this using the Pythagorean theorem. The hypotenuse RT has length

12 + 27 = 39

. If we call the legs of the original triangle

a

and

b

, then

a^2 = 12 \times 39 = 468

and

b^2 = 27 \times 39 = 1053

. Indeed,

468 + 1053 = 1521 = 39^2

. Answer choice A (15) would result from incorrectly taking the arithmetic mean:

\frac{12 + 27}{2} = 19.5

, then rounding. Answer choice B (24) might come from doubling the shorter segment. Answer choice C (20) could result from various computational errors or misapplying formulas. Remember: when an altitude is drawn to the hypotenuse of a right triangle, always use the geometric mean formula

h = \sqrt{segment_1 \times segment_2}

. This relationship appears frequently on geometry problems.

Question 3

Triangle PQR is a right triangle with right angle at Q. The altitude from Q meets hypotenuse PR at point S, creating segments PS = 9 and SR = 16. If triangle PQS is similar to triangle QRS, what is the length of the longer leg of triangle PQR?

$15$
$20$ (correct answer)
$12$
$25$

Explanation: First, find QS using the geometric mean: QS = √(9 × 16) = 12. The legs PQ and QR can be found using the relationships: PQ² = PS × PR = 9 × 25 = 225, so PQ = 15. QR² = SR × PR = 16 × 25 = 400, so QR = 20. The longer leg is 20. Choice A is the shorter leg. Choice C is the altitude length. Choice D is the hypotenuse length.

Question 4

Triangle PQR is a right triangle with right angle at Q. The altitude from Q to hypotenuse PR has length 12, and one of the segments of the hypotenuse has length 16. What is the length of the other segment of the hypotenuse?

$9$ (correct answer)
$8$
$6$
$10$

Explanation: Using the geometric mean relationship in right triangles, the altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse. If the segments are 16 and s, then 12² = 16s, so 144 = 16s, giving s = 9. Choice B results from incorrectly using 12 = 16 - s. Choice C comes from using 12 = √(16 + s). Choice D assumes the segments differ by 6.

Question 5

In right triangle MNP with right angle at N, the altitude from N to hypotenuse MP creates point Q on MP. If MQ = 7, QP = 28, and triangle MNQ is similar to triangle PNM, what is the length of NQ?

$12$
$21$
$35$
$14$ (correct answer)

Explanation: When you see an altitude drawn to the hypotenuse of a right triangle, you're dealing with geometric mean relationships. This configuration creates three similar triangles, and the altitude's length is the geometric mean of the segments it creates on the hypotenuse. In this problem, altitude NQ divides hypotenuse MP into segments MQ = 7 and QP = 28. The key relationship is:

NQ^2 = MQ \times QP

. This comes from the similarity of the three triangles formed. Substituting the given values:

NQ^2 = 7 \times 28 = 196

. Therefore,

NQ = \sqrt{196} = 14

. Looking at the wrong answers: Choice A (

12

) might come from incorrectly trying to find some average or using the Pythagorean theorem inappropriately. Choice B (

21

) is the arithmetic mean of 7 and 28, which is a common mistake—students sometimes confuse arithmetic and geometric means. Choice C (

35

) is simply the sum of the two segments, which has no geometric significance in this context. The confirmation that triangle MNQ is similar to triangle PNM tells you that the geometric mean relationship applies, but you don't need to use the similarity directly—the altitude-to-hypotenuse theorem gives you the answer immediately. Remember: whenever you see an altitude drawn to the hypotenuse of a right triangle, think geometric mean. The altitude squared equals the product of the two segments it creates on the hypotenuse.

Question 6

Two right triangles are similar with a ratio of corresponding sides of 3:4. If the altitude to the hypotenuse in the smaller triangle has length 7.2, what is the length of the altitude to the hypotenuse in the larger triangle?

$10.8$
$9.6$ (correct answer)
$12$
$8.4$

Explanation: When you encounter similar triangles, remember that ALL corresponding linear measurements scale by the same ratio. This includes not just the obvious sides, but also altitudes, medians, angle bisectors, and any other linear distances within the triangles. Since these triangles are similar with a ratio of 3:4 (smaller to larger), every linear measurement in the larger triangle will be

\frac{4}{3}

times the corresponding measurement in the smaller triangle. The altitude to the hypotenuse in the smaller triangle is 7.2, so the altitude to the hypotenuse in the larger triangle must be

7.2 \times \frac{4}{3} = 7.2 \times \frac{4}{3} = \frac{28.8}{3} = 9.6

. Looking at the wrong answers: Choice A (

10.8

) represents multiplying by

\frac{3}{2}

instead of

\frac{4}{3}

, which confuses the ratio. Choice C (

12

) appears to use an incorrect scaling factor, possibly treating this as a 2:3 ratio problem. Choice D (

8.4

) represents adding a constant difference rather than using proportional scaling, which is a common error when students forget that similarity requires multiplicative relationships, not additive ones. The key insight is that similarity preserves all proportional relationships within the figures. When triangles are similar with ratio

a:b

, every corresponding linear measurement scales by exactly that same ratio. Don't let the specific type of measurement (altitude, side, median, etc.) confuse you—the scaling factor remains constant for all linear dimensions.

Question 7

In right triangle ABC with altitude h drawn to hypotenuse c, if the two segments of the hypotenuse are in the ratio 1:4 and the altitude has length 6, what is the length of the shorter leg of the original triangle?

$3\sqrt{5}$ (correct answer)
$6\sqrt{5}$
9
18

Explanation: Let the segments be x and 4x. Using h² = (segment 1)(segment 2): 36 = x(4x) = 4x², so x = 3. The segments are 3 and 12. For the shorter leg, using leg² = (adjacent segment)(whole hypotenuse): leg² = 3(15) = 45, so leg = √45 = 3√5. Choice B uses the longer leg. Choice C is just one segment length. Choice D is the total hypotenuse length.

Question 8

Two similar right triangles have altitudes to their hypotenuses in the ratio 3:5. If the smaller triangle has a hypotenuse of length 15 and the larger triangle has an area of 150, what is the area of the smaller triangle?

54 (correct answer)
90
96
135

Explanation: If the altitudes are in ratio 3:5, then the triangles are similar with ratio 3:5, so their areas are in ratio 9:25. Let the smaller triangle's area be A. Then A/150 = 9/25, so A = 150 × 9/25 = 54. Choice B assumes a 3:5 area ratio instead of 9:25. Choice C uses incorrect altitude relationships. Choice D assumes the linear ratio applies to areas.

Question 9

Triangle DEF is similar to triangle GHI with a ratio of similarity of 2:3. If triangle DEF has sides 8, 15, and 17, and triangle GHI is a right triangle, which similarity relationship involving altitudes is correct?

The altitude to the hypotenuse in DEF is $\frac{2}{3}$ times the corresponding altitude in GHI (correct answer)
The altitude to the hypotenuse in GHI is $\frac{2}{3}$ times the corresponding altitude in DEF
The ratio of altitudes equals the ratio of areas, which is $\frac{4}{9}$
The altitude to the hypotenuse in DEF is $\frac{4}{9}$ times the corresponding altitude in GHI

Explanation: Since DEF ~ GHI with ratio 2:3, all corresponding linear measurements (including altitudes) have the same ratio. The altitude in DEF is 2/3 times the corresponding altitude in GHI. Choice B reverses the ratio. Choice C confuses linear ratio with area ratio. Choice D incorrectly uses the area ratio for linear measurements.

Question 10

In right triangle ABC with right angle at C, altitude CD is drawn to hypotenuse AB. If AD = x, DB = 4x, and CD = 8, what is the value of x?

$6$
$2$
$4$ (correct answer)
$3$

Explanation: When you encounter a right triangle with an altitude drawn to the hypotenuse, you're dealing with geometric mean relationships. This setup creates three similar triangles with powerful proportional relationships. In this problem, altitude CD creates two segments on hypotenuse AB: AD = x and DB = 4x. The key relationship is that the altitude to the hypotenuse equals the geometric mean of the two segments it creates. This gives us the equation:

CD^2 = AD \times DB

Substituting the known values:

8^2 = x \times 4x

64 = 4x^2

x^2 = 16

x = 4

Let's check why the other answers don't work. Choice A (

x = 6

) would give us

4x^2 = 4(36) = 144

, but

CD^2 = 64

, so this is too large. Choice B (

x = 2

) yields

4x^2 = 4(4) = 16

, which is too small since we need

CD^2 = 64

. Choice D (

x = 3

) produces

4x^2 = 4(9) = 36

, still short of our required 64. The answer is C:

x = 4

. Study tip: Memorize the altitude-to-hypotenuse theorem: when an altitude is drawn to the hypotenuse of a right triangle, the altitude squared equals the product of the two segments of the hypotenuse. This relationship appears frequently on geometry exams and provides a direct path to the solution.

Question 11

In similar right triangles ABC and DEF, the ratio of their corresponding altitudes to their hypotenuses is 2:5. If the hypotenuse of triangle ABC is 26, what is the hypotenuse of triangle DEF?

$39$
$52$
$65$ (correct answer)
$58.5$

Explanation: When you encounter problems involving similar triangles, remember that all corresponding measurements maintain the same ratio - sides, altitudes, perimeters, and any other linear dimensions are proportional. Since triangles ABC and DEF are similar with a 2:5 ratio for corresponding altitudes to hypotenuses, this same 2:5 ratio applies to the hypotenuses themselves. If triangle ABC has hypotenuse 26, then you can set up the proportion:

\frac{\text{hypotenuse of ABC}}{\text{hypotenuse of DEF}} = \frac{2}{5}

Substituting:

\frac{26}{\text{hypotenuse of DEF}} = \frac{2}{5}

Cross-multiplying:

2 \times \text{hypotenuse of DEF} = 26 \times 5 = 130

Therefore:

\text{hypotenuse of DEF} = \frac{130}{2} = 65

Looking at the wrong answers: Choice (A) 39 results from incorrectly thinking the ratio should be

\frac{26}{39} = \frac{2}{3}

, confusing the given ratio. Choice (B) 52 comes from simply doubling the given hypotenuse, ignoring the ratio entirely. Choice (D) 58.5 appears to result from using

\frac{26}{x} = \frac{4}{9}

or similar incorrect ratio manipulation. The answer is (C) 65. Study tip: In similar triangle problems, identify what ratio is given first, then remember that ALL linear measurements share that same ratio. Set up your proportion carefully, making sure the corresponding parts are in the same positions in your fraction.