Maths · Geometry · chapter 6 of 14

Class 10 Maths Chapter 6: Triangles worksheet with answer key

A 20-mark worksheet on Triangles for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Download the worksheet (PDF)Answer key (PDF)
Question 1
A photograph 10 cm × 15 cm is enlarged to a print 20 cm × 30 cm. The photograph and the enlarged print are
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Answer: (b) similar but not congruent

Both are rectangles, so all angles are 90°, and 20/10 = 30/15 = 2: same shape, different size. Similar, not congruent.

Question 2
Which of the following statements is always true?
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Answer: (b) Any two congruent triangles are similar.

Congruent triangles have equal angles and sides in the ratio 1 : 1, so they are similar; similar triangles need not be the same size.

Question 3
On a map of a town, 1 cm stands for 500 m. The map and the town are similar figures, with representative fraction
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Answer: (d) 150000

500 m = 50 000 cm, so 1 cm on the map stands for 50 000 cm on the ground: the scale factor is 1/50 000.

Question 4
In ΔABC, ∠A = 52° and ∠B = 71°. In ΔXYZ, ∠X = 52° and ∠Y = 57°. Then
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Answer: (d) ΔABC ~ ΔXZY

∠C = 180° − 52° − 71° = 57° and ∠Z = 180° − 52° − 57° = 71°. So A ↔ X, B ↔ Z, C ↔ Y, and ΔABC ~ ΔXZY by AA.

Question 5
Which pair of figures is NOT necessarily similar?
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Answer: (a) two rhombuses, each of side 5 cm

Regular hexagons have every angle 120° and proportional sides; all circles are similar; a right isosceles triangle always has angles 90°, 45°, 45°. Two rhombuses of side 5 cm have equal sides but their angles can differ, so they need not be similar.

Question 6 · 2 marks
D and E lie on sides AB and AC of ΔABC with AD = x cm, DB = (x + 3) cm, AE = (x + 2) cm and EC = (x + 8) cm. Find x for which DE ∥ BC.
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DE ∥ BC exactly when AD/DB = AE/EC: x/(x + 3) = (x + 2)/(x + 8). Then x² + 8x = x² + 5x + 6, so 3x = 6 and x = 2. (Check: 2/5 = 4/10.)

  • sets up x/(x + 3) = (x + 2)/(x + 8) (1)
  • x = 2 (1)
Question 7 · 2 marks
A rectangle measures 8 cm by 12 cm. A parallelogram has sides 4 cm and 6 cm and one angle of 60°. Are the two figures similar? Give a reason.
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Sides: 8/4 = 12/6 = 2, so they are in the same ratio. Angles: the rectangle's are 90°, the parallelogram has a 60° angle, so corresponding angles are not equal. Both conditions are needed, so the figures are not similar.

  • sides in the same ratio, 2 (1)
  • angles 90° and 60° differ, so not similar (1)
Question 8 · 2 marks
ΔDEF is right-angled at E with DF = 17 cm and DE = 8 cm. ΔLMN is right-angled at M with LN = 25.5 cm and LM = 12 cm. Show that the triangles are similar and find MN.
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LN/DF = 25.5/17 = 1.5 and LM/DE = 12/8 = 1.5: the hypotenuse and one side are proportional, so the right triangles are similar (RHS similarity), ΔDEF ~ ΔLMN. EF = √(17² − 8²) = 15 cm, so MN = 1.5 × 15 = 22.5 cm.

  • hypotenuse and side in ratio 1.5, so similar (1)
  • MN = 22.5 cm (1)
Question 9 · 3 marks
In ΔPQR, S lies on PQ and T on PR such that PS·PQ = PT·PR. PS = 4 cm, PQ = 15 cm, PT = 5 cm and PR = 12 cm. (i) Prove that ΔPST ~ ΔPRQ. (ii) If QR = 10.5 cm, find ST.
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(i) PS·PQ = PT·PR (4 × 15 = 5 × 12 = 60) can be written PS/PR = PT/PQ, both 1/3. ∠SPT = ∠RPQ (common). The sides including the equal angles are proportional, so ΔPST ~ ΔPRQ by SAS, with S ↔ R and T ↔ Q. (ii) ST/RQ = PS/PR = 1/3, so ST = 10.5/3 = 3.5 cm.

  • rewrites the product as PS/PR = PT/PQ (1)
  • SAS with the common angle, ΔPST ~ ΔPRQ (1)
  • ST = 3.5 cm (1)
Question 10 · 3 marks
In ΔABC, P and Q lie on AB and AC with AP = 3.6 cm, PB = 5.4 cm, AQ = 2.4 cm and QC = 3.6 cm. (i) Show that PQ ∥ BC. (ii) If BC = 7.5 cm, find PQ.
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(i) AP/PB = 3.6/5.4 = 2/3 and AQ/QC = 2.4/3.6 = 2/3. The ratios are equal, so PQ ∥ BC by the converse of BPT. (ii) With PQ ∥ BC, ∠APQ = ∠ABC and ∠AQP = ∠ACB (corresponding angles), so ΔAPQ ~ ΔABC (AA). PQ/BC = AP/AB = 3.6/9 = 2/5, so PQ = 7.5 × 2/5 = 3 cm.

  • both ratios 2/3, PQ ∥ BC by the converse (1)
  • ΔAPQ ~ ΔABC by AA (1)
  • PQ = 3 cm (1)
Question 11 · 3 marks
In ΔABC, AB = AC. Points D and E lie on AB and AC such that AD = AE. Prove that DE ∥ BC.
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DB = AB − AD and EC = AC − AE. Since AB = AC and AD = AE, DB = EC. Hence AD/DB = AE/EC. A line dividing two sides of a triangle in the same ratio is parallel to the third side, so DE ∥ BC.

  • DB = EC by subtraction (1)
  • AD/DB = AE/EC (1)
  • DE ∥ BC by the converse of BPT (1)
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