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ACT Math

ACT Math Help: Similarity And Congruence

Review real example questions for Similarity And Congruence in ACT Math.

Question 1

In the diagram, △ABC∼△DEF\triangle ABC \sim \triangle DEF△ABC∼△DEF with vertex correspondence A↔DA\leftrightarrow DA↔D, B↔EB\leftrightarrow EB↔E, C↔FC\leftrightarrow FC↔F. Side lengths shown are AB6ˉAB\=6AB6ˉ, BC1ˉ0BC\=10BC1ˉ0, DE9ˉDE\=9DE9ˉ, and EFxˉEF\=xEFxˉ.

What is the length of side xxx using similarity?

  1. x=12x=12x=12
  2. x=15x=15x=15
  3. x=203x=\dfrac{20}{3}x=320​
  4. x=275x=\dfrac{27}{5}x=527​
Explanation: Triangles ABC and DEF are similar with given vertex correspondences A to D, B to E, and C to F. Corresponding sides include AB to DE and BC to EF. The ratio of corresponding sides is DE/AB = 9/6 = 3/2, establishing the scale factor from ABC to DEF. Applying this to BC and EF gives EF = 10 × 3/2 = 15, so x = 15. This proportion ensures consistency across corresponding sides and avoids misapplying the scale factor direction.

Question 2

Two triangles are shown with markings indicating equal parts. In △ABC\triangle ABC△ABC and △DEF\triangle DEF△DEF, ∠A\angle A∠A is marked congruent to ∠D\angle D∠D (one arc), and ∠B\angle B∠B is marked congruent to ∠E\angle E∠E (two arcs). The side between those angles, ABABAB, has one tick mark, and the corresponding side DEDEDE also has one tick mark.

Which congruence criterion applies (SSS, SAS, ASA, AAS)?

  1. ASA
  2. AAS
  3. SAS
  4. SSS
Explanation: Triangles ABC and DEF are congruent by ASA congruence because two pairs of corresponding angles are equal and the included sides are equal. The correspondences are angle A to angle D (one arc) and angle B to angle E (two arcs), with included side AB to DE (one tick each). With the equal angles surrounding the equal included side, all corresponding parts are equal. This distinguishes ASA from AAS, which involves a non-included side, emphasizing the importance of the side's position.

Question 3

Two triangles are shown. In △ABC\triangle ABC△ABC, AB=6AB=6AB=6, AC=9AC=9AC=9, BC=12BC=12BC=12. In △DEF\triangle DEF△DEF, DE=4DE=4DE=4, DF=6DF=6DF=6, EF=8EF=8EF=8. What is the scale factor from △ABC\triangle ABC△ABC to △DEF\triangle DEF△DEF (i.e., multiply lengths in △ABC\triangle ABC△ABC by what number to get corresponding lengths in △DEF\triangle DEF△DEF)?

  1. 32\dfrac{3}{2}23​
  2. 23\dfrac{2}{3}32​
  3. 43\dfrac{4}{3}34​
  4. 12\dfrac{1}{2}21​
Explanation: To find the scale factor from triangle ABC to triangle DEF, we need to check if the triangles are similar by comparing ratios of corresponding sides. Let's check: DE/AB = 4/6 = 2/3, DF/AC = 6/9 = 2/3, and EF/BC = 8/12 = 2/3. Since all three ratios are equal, the triangles are similar by SSS similarity. The scale factor from triangle ABC to triangle DEF is 2/3, meaning we multiply each side length in triangle ABC by 2/3 to get the corresponding side length in triangle DEF.

Question 4

In △ABC\triangle ABC△ABC and △DEF\triangle DEF△DEF, ∠A≅∠D\angle A\cong\angle D∠A≅∠D, ∠B≅∠E\angle B\cong\angle E∠B≅∠E, and the side between them is equal: AB=DE=7AB=DE=7AB=DE=7. Which congruence criterion applies (SSS, SAS, ASA, AAS)?

  1. AAS
  2. ASA
  3. SAS
  4. SSS
Explanation: The triangles are congruent by ASA (Angle-Side-Angle) criterion. We have angle A ≅ angle D, angle B ≅ angle E, and the side between these angles is equal: AB = DE = 7. In ASA congruence, we need two angles and the included side (the side between the two angles) to be equal. Since AB is the side between angles A and B, and DE is the side between angles D and E, and these sides are equal along with their adjacent angles, the triangles are congruent by ASA.

Question 5

Triangles △JKL\triangle JKL△JKL and △MNO\triangle MNO△MNO are similar. Corresponding sides are JK↔MNJK \leftrightarrow MNJK↔MN, KL↔NOKL \leftrightarrow NOKL↔NO, and JL↔MOJL \leftrightarrow MOJL↔MO. If JK=8JK=8JK=8, KL=10KL=10KL=10, JL=12JL=12JL=12, and MN=12MN=12MN=12, what is the length of NONONO?

  1. 121212
  2. 151515
  3. 203\dfrac{20}{3}320​
  4. 253\dfrac{25}{3}325​
Explanation: The triangles are similar with given correspondences: JK ↔ MN, KL ↔ NO, and JL ↔ MO. First, find the scale factor using the known corresponding sides: MN/JK = 12/8 = 3/2. Since the triangles are similar, all corresponding sides have the same ratio. To find NO, we use the proportion: NO/KL = 3/2. Therefore, NO = KL × (3/2) = 10 × (3/2) = 15. The length of NO is 15.

Question 6

Triangles △GHI\triangle GHI△GHI and △JKL\triangle JKL△JKL are similar by AA with correspondence G↔JG\leftrightarrow JG↔J, H↔KH\leftrightarrow KH↔K, I↔LI\leftrightarrow LI↔L. If GH=12GH=12GH=12, JK=8JK=8JK=8, and HI=15HI=15HI=15, what is the length of the corresponding side KLKLKL?

  1. 101010
  2. 181818
  3. 202020
  4. 22.522.522.5
Explanation: The triangles are similar by AA with G↔J, H↔K, I↔L, so HI corresponds to KL. The scale factor from △GHI to △JKL is JK/GH = 8/12 = 2/3. Therefore, KL = HI × scale factor = 15 × (2/3) = 10. Note that we're scaling down from the larger to the smaller triangle, so we multiply by 2/3.

Question 7

Triangles △PQR\triangle PQR△PQR and △STU\triangle STU△STU are similar by AA. Angle ∠P\angle P∠P corresponds to ∠S\angle S∠S, and ∠Q\angle Q∠Q corresponds to ∠T\angle T∠T. If PQ=6PQ=6PQ=6 and the corresponding side ST=9ST=9ST=9, what is the scale factor from △PQR\triangle PQR△PQR to △STU\triangle STU△STU?

  1. 23\dfrac{2}{3}32​
  2. 32\dfrac{3}{2}23​
  3. 53\dfrac{5}{3}35​
  4. 35\dfrac{3}{5}53​
Explanation: The triangles are similar by AA, with P↔S and Q↔T, so side PQ corresponds to side ST. The scale factor from △PQR to △STU is the ratio of corresponding sides: ST/PQ = 9/6 = 3/2. This means each side of △STU is 3/2 times the corresponding side of △PQR. The scale factor is 3/2, not 2/3, because we're scaling from the smaller to the larger triangle.

Question 8

In triangle ABC and triangle XYZ, if angle A=angle X\text{angle } A = \text{angle } Xangle A=angle X and angle B=angle Y\text{angle } B = \text{angle } Yangle B=angle Y, are the triangles similar?

  1. Yes, by AA
  2. No
  3. Yes, by SSS
  4. Yes, by SAS
Explanation: The triangles are similar by AA (Angle-Angle) criterion. We have two pairs of equal angles: angle A = angle X and angle B = angle Y. When two angles of one triangle are equal to two angles of another triangle, the third angles must also be equal by the angle sum property. This is sufficient to prove similarity using the AA criterion.

Question 9

For triangles △XYZ\triangle XYZ△XYZ and △ABC\triangle ABC△ABC, △XYZ\triangle XYZ△XYZ has angles X=45oX = 45^\text{o}X=45o, Y=45oY = 45^\text{o}Y=45o, and △ABC\triangle ABC△ABC has angles A=45oA = 45^\text{o}A=45o, B=45oB = 45^\text{o}B=45o. Are the triangles similar?

  1. Yes, by SSS similarity.
  2. No, they are not similar.
  3. Yes, by SAS similarity.
  4. Yes, by AA similarity.
Explanation: The triangles are similar by AA similarity criterion because they have two pairs of equal angles. Triangle XYZ has angles 45°, 45°, and 90° (since angles sum to 180°). Triangle ABC has angles 45°, 45°, and 90° (since angles sum to 180°). Having two pairs of equal angles (45° = 45° and 45° = 45°) confirms similarity by AA criterion.

Question 10

Two triangles, △GHI\triangle GHI△GHI and △JKL\triangle JKL△JKL, △GHI\triangle GHI△GHI has angles G=60oG = 60^\text{o}G=60o, H=60oH = 60^\text{o}H=60o, and △JKL\triangle JKL△JKL has angles J=60oJ = 60^\text{o}J=60o, K=60oK = 60^\text{o}K=60o. Are the triangles similar?

  1. Yes, by SSS similarity.
  2. No, they are not similar.
  3. Yes, by AA similarity.
  4. Yes, by SAS similarity.
Explanation: The triangles are similar by AA similarity criterion because they have two pairs of equal angles. Triangle GHI has angles 60°, 60°, and 60° (since it's equilateral with angles summing to 180°). Triangle JKL has angles 60°, 60°, and 60° (since it's equilateral with angles summing to 180°). Having two pairs of equal angles (60° = 60° and 60° = 60°) confirms similarity by AA criterion.