Chapter 6: Triangles — Online MCQ Test
MATHS · CLASS 10 · Andhra State Board
Practice Chapter 6: Triangles with a free chapter-wise online MCQ test.
This chapter covers: Similar triangles Criteria for similarity Basic proportionality theorem Pythagoras theorem Converse of Pythagoras theorem Triangle proofs Ratio of areas.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter 6: Triangles — Important Questions & Answers
Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are:
- A. Equal
- B. Perpendicular
- C. Proportional
- D. Parallel
Answer: C. Proportional
Similar triangles have equal corresponding angles and proportional corresponding sides.
Similar triangles have equal corresponding angles and proportional corresponding sides.
Which similarity criterion states that two triangles are similar if their corresponding angles are equal?
- A. SSS similarity
- B. AAA similarity
- C. RHS congruence
- D. SAS congruence
Answer: B. AAA similarity
By AAA similarity, if all corresponding angles of two triangles are equal, the triangles are similar.
By AAA similarity, if all corresponding angles of two triangles are equal, the triangles are similar.
In triangle ABC, D lies on AB and E lies on AC. If DE is parallel to BC, AD = 3 cm, DB = 4 cm and AE = 6 cm, then EC is:
- A. 6 cm
- B. 7 cm
- C. 8 cm
- D. 10 cm
Answer: C. 8 cm
By BPT, AD/DB = AE/EC. So 3/4 = 6/EC, giving EC = 8 cm.
By BPT, AD/DB = AE/EC. So 3/4 = 6/EC, giving EC = 8 cm.
Two vertical poles cast shadows at the same time. A 6 m pole casts a 4 m shadow. If another pole casts an 18 m shadow, its height is:
- A. 12 m
- B. 18 m
- C. 24 m
- D. 27 m
Answer: D. 27 m
The pole and its shadow form similar right triangles. Therefore height/shadow = 6/4, so height = 18 × 6/4 = 27 m.
The pole and its shadow form similar right triangles. Therefore height/shadow = 6/4, so height = 18 × 6/4 = 27 m.
In triangle ABC, P lies on AB and Q lies on AC such that PQ is parallel to BC. If AP:PB = 2:3 and the area of trapezium PBCQ is 84 cm², then the area of triangle APQ is:
- A. 12 cm²
- B. 16 cm²
- C. 20 cm²
- D. 24 cm²
Answer: B. 16 cm²
AP/AB = 2/5, so ar(APQ)/ar(ABC) = 4/25. The trapezium area is 21/25 of ar(ABC), so ar(ABC) = 100 cm² and ar(APQ) = 16 cm².
AP/AB = 2/5, so ar(APQ)/ar(ABC) = 4/25. The trapezium area is 21/25 of ar(ABC), so ar(ABC) = 100 cm² and ar(APQ) = 16 cm².