Math 2 Quiz: Using Corresponding Parts
7 questions · exam conditions
0:00
Using Corresponding PartsQuestion 1 of 7

In right triangle RST with right angle at S, the altitude from S to hypotenuse RT creates two smaller triangles. If these smaller triangles have areas of 24 square units and 54 square units respectively, what is the length of the altitude from S to RT?

12
15
18
24
← Back to quizzes

Math 2 Quiz

Math 2 Quiz: Using Corresponding Parts

Practice Using Corresponding Parts 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 Using Corresponding Parts, 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

In right triangle RST with right angle at S, the altitude from S to hypotenuse RT creates two smaller triangles. If these smaller triangles have areas of 24 square units and 54 square units respectively, what is the length of the altitude from S to RT?

  1. 12 (correct answer)
  2. 15
  3. 18
  4. 24
Explanation: Let the altitude from S to RT have length h, and let it meet RT at point U, creating segments RU = a and UT = b. The two smaller triangles RSU and STU have areas 24 and 54 respectively. So (1/2) × a × h = 24, giving ah = 48, and (1/2) × b × h = 54, giving bh = 108. From the geometric mean relationship in right triangles with altitude to the hypotenuse, h² = ab. Substituting a = 48/h and b = 108/h: h² = (48/h)(108/h) = 5184/h². Therefore h⁴ = 5184, and h = ⁴√5184 = 12.

Question 2

Triangles JKL and MNP are similar with JK corresponding to MN, KL corresponding to NP, and LJ corresponding to PM. If JK = 12, MN = 18, and the difference between corresponding altitudes from K and N is 4 units, what is the altitude from K to side JL?

  1. 8 units (correct answer)
  2. 6 units
  3. 10 units
  4. 12 units
Explanation: The scale factor from triangle JKL to triangle MNP is MN/JK = 18/12 = 3/2. Let h₁ be the altitude from K in triangle JKL and h₂ be the corresponding altitude from N in triangle MNP. Since altitudes are proportional to the scale factor, h₂ = h₁ × (3/2). Given that h₂ - h₁ = 4, we have h₁ × (3/2) - h₁ = 4, which gives h₁ × (1/2) = 4, so h₁ = 8 units.

Question 3

In right triangle ABC with right angle at C, point D is on hypotenuse AB such that CD ⊥ AB. If AD = 9 and DB = 16, what is the length of CD?

  1. 12 (correct answer)
  2. 12.5
  3. 15
  4. 25
Explanation: When an altitude is drawn to the hypotenuse of a right triangle, it creates two smaller triangles that are similar to the original triangle and to each other. In this configuration, CD² = AD × DB by the geometric mean relationship. Therefore, CD² = 9 × 16 = 144, so CD = 12. Choice B uses the arithmetic mean (9+16)/2. Choice C incorrectly uses CD = √(AD² + DB²)/2. Choice D uses AD + DB.

Question 4

Two similar triangles have corresponding altitudes in the ratio 5:7. If the area of the smaller triangle is 75 square units, what is the area of the larger triangle?

  1. 105 square units
  2. 147 square units (correct answer)
  3. 210 square units
  4. 245 square units
Explanation: If corresponding altitudes are in the ratio 5:7, then all corresponding linear measurements are in this ratio since the triangles are similar. The ratio of areas equals the square of the ratio of corresponding linear measurements: (7/5)² = 49/25. Therefore, the area of the larger triangle is 75 × (49/25) = 147. Choice A uses the linear ratio 7/5 instead of squaring it. Choice C uses an incorrect calculation. Choice D uses an arithmetic error.

Question 5

In triangle ABC, the angle bisector of angle A intersects side BC at point D. If AB = 15, AC = 20, and BD = 9, what is the length of DC?

  1. 12 (correct answer)
  2. 15
  3. 18
  4. 21
Explanation: By the Angle Bisector Theorem, when an angle bisector divides the opposite side, it creates segments proportional to the adjacent sides. Therefore, BD/DC = AB/AC. Substituting: 9/DC = 15/20 = 3/4. Cross-multiplying: 9 × 4 = 3 × DC, so 36 = 3 × DC, giving DC = 12. Choice B assumes DC = AB. Choice C uses the incorrect ratio 9/DC = 20/15. Choice D uses 9/DC = 15/20 but solves incorrectly as DC = 9 × 20/15.

Question 6

In triangle ABC, point D lies on side AB and point E lies on side AC such that DE is parallel to BC. If AD = 6, DB = 9, and AE = 8, what is the length of EC?

  1. 15
  2. 10
  3. 12 (correct answer)
  4. 18
Explanation: When you see a triangle with a line segment parallel to one of its sides, you're dealing with similar triangles and proportional segments. This is a fundamental property: if a line is drawn parallel to one side of a triangle, it creates two similar triangles. Since DE is parallel to BC, triangles ADE and ABC are similar. This means their corresponding sides are proportional. The key relationship is: ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC} First, find the total length AB: AD + DB = 6 + 9 = 15. Let EC = x, so AC = AE + EC = 8 + x. Setting up the proportion: 615=88+x\frac{6}{15} = \frac{8}{8 + x} Cross-multiplying: 6(8+x)=15×86(8 + x) = 15 \times 8 48+6x=12048 + 6x = 120 6x=726x = 72 x=12x = 12 Therefore, EC = 12, which is answer C. Looking at the wrong answers: A) 15 incorrectly assumes EC equals the total length AB. B) 10 might result from setting up an incorrect proportion or arithmetic error. D) 18 could come from incorrectly using the ratio DBAD=96=32\frac{DB}{AD} = \frac{9}{6} = \frac{3}{2} and multiplying AE by this ratio (8 × 3/2 = 12, but then adding 6 to get 18). Study tip: When you see parallel lines in triangles, immediately think "similar triangles and proportional segments." Set up your proportion carefully, ensuring corresponding sides are in the same position in both ratios. Double-check by verifying that your answer maintains the proportional relationship.

Question 7

In triangle XYZ, point M is on side XY and point N is on side XZ such that MN || YZ. If XM = 4, MY = 12, and XN = 5, then what is the ratio of the area of triangle XMN to the area of triangle XYZ?

  1. 4:25
  2. 1:9
  3. 1:16 (correct answer)
  4. 1:4
Explanation: When you see parallel lines in triangles, you're dealing with similar triangles and proportional relationships. Since MN || YZ, triangle XMN is similar to triangle XYZ, which means their corresponding sides are proportional. First, find the ratio of corresponding sides. Since XM = 4 and MY = 12, the total length XY = 4 + 12 = 16. The ratio of similarity is XMXY=416=14\frac{XM}{XY} = \frac{4}{16} = \frac{1}{4}. You can verify this with the other side: since the triangles are similar, XNXZ=14\frac{XN}{XZ} = \frac{1}{4}, so XZ = 4 × 5 = 20. For similar triangles, the ratio of their areas equals the square of the ratio of corresponding sides. Since the side ratio is 14\frac{1}{4}, the area ratio is (14)2=116\left(\frac{1}{4}\right)^2 = \frac{1}{16}. Therefore, the ratio is 1:16. Let's examine why the other answers are wrong. Choice A (4:25) incorrectly uses 420=15\frac{4}{20} = \frac{1}{5} as the area ratio instead of squaring it to get 125\frac{1}{25}, then flips the ratio. Choice B (1:9) mistakenly uses the ratio 412=13\frac{4}{12} = \frac{1}{3} (comparing XM to MY instead of XM to XY) and squares it. Choice D (1:4) uses the correct side ratio but forgets to square it for the area ratio. Remember: when dealing with similar figures, always square the side ratio to get the area ratio. This relationship appears frequently on geometry problems involving parallel lines in triangles.