Triangles CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Triangles CBQs

1. Question 1MCQ1 mark
   The roof of a shed is supported by a triangular wooden truss ABC, where AB and AC are the two sloping rafters and BC is the base beam. A horizontal tie-beam DE is fixed parallel to BC, with D on rafter AB and E on rafter AC. The carpenter measures AD = 1.8 m, DB = 2.7 m and AE = 2 m. The length of EC is
   • 3 m
   • 1.33 m
   • 2.43 m
   • 4.5 m
   Show step-wise solution

   1. [1 mark] DE ∥ BC, so by the Basic Proportionality Theorem AD/DB = AE/EC ⇒ 1.8/2.7 = 2/EC ⇒ EC = (2 × 2.7)/1.8 = 3 m.

   Answer: EC = 3 m

2. Question 2Assertion-Reason1 mark
   A triangular tent flap ABC has a straight zip DE sewn with D on edge AB and E on edge AC. The tailor measures AD = 1.2 m, DB = 1.8 m, AE = 1.6 m and EC = 2.4 m. Assertion (A): The zip DE is parallel to the bottom edge BC. Reason (R): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): AD/DB = 1.2/1.8 = 2/3 and AE/EC = 1.6/2.4 = 2/3, so by the converse of BPT (statement R) DE ∥ BC. Both are true and R explains A.

   Answer: (A)

3. Question 3Short answer2 marks
   At 4 p.m. on a sunny day, Meera, who is 1.6 m tall, stands upright on level ground and casts a shadow 2 m long. At the same moment, a vertical electric pole standing on the same level ground casts a shadow 9.5 m long. Find the height of the pole.
   Show step-wise solution

   1. [1 mark] The Sun's rays are parallel and both Meera and the pole are vertical, so the two right triangles (height, shadow, ray) are similar by AA: h/9.5 = 1.6/2.
   2. [1 mark] h = 9.5 × 0.8 = 7.6 m.

   Answer: Height of the pole = 7.6 m

4. Question 4Case study4 marks
   Meera makes a pinhole camera from a closed box 20 cm long. Light from a tree passes through a pinhole P in the front face and forms an inverted image on a translucent screen that forms the back face. The tree and the screen are both vertical. The foot F of the tree, the pinhole P and the point F′ on the screen where the image of the foot is formed lie on one horizontal line that is perpendicular to both the tree and the screen. The tree is 30 m from the pinhole (PF = 30 m), PF′ = 20 cm, and the image of the tree on the screen is 4 cm tall. Based on the above information, answer the following questions: (i) Name the similarity criterion by which the triangle formed by P with the tree is similar to the triangle formed by P with the image. (ii) Find the ratio PF : PF′ (in the same units). (iii) Find the height of the tree in metres.
   Show step-wise solution

   1. [1 mark] (i) The angles at F and F′ are both 90°, and the two angles at P are vertically opposite, hence equal. So the triangles are similar by the AA criterion.
   2. [1 mark] (ii) PF = 30 m = 3000 cm, so PF : PF′ = 3000 : 20 = 150 : 1.
   3. [1 mark] (iii) Corresponding sides of the similar triangles are proportional: height of tree ÷ height of image = PF ÷ PF′ = 150, so height of tree = 150 × 4 cm = 600 cm.
   4. [1 mark] (iii) Height of the tree = 600 cm = 6 m.

   Answer: (i) AA criterion (ii) PF : PF′ = 150 : 1 (iii) height of tree = 6 m

5. Question 5Long answer5 marks
   At night a boy 1.5 m tall stands on level ground at a distance of d m from the foot of a vertical street-lamp pole of height h m, and his shadow on the ground is 2 m long. He then walks 3 m directly away from the pole along the same straight line, and his shadow becomes 3 m long. Treat the lamp as a point at the top of the pole. (a) Using similar triangles, write two equations connecting h and d. (b) Find d and the height h of the lamp. (c) How long will his shadow be when he stands 15 m from the foot of the pole?
   Show step-wise solution

   1. [1 mark] (a) The pole and the boy are both vertical, and the triangle formed by the boy and his shadow shares the angle at the shadow tip with the triangle formed by the pole and the distance to the tip, so the two triangles are similar (AA). First position: shadow tip is d + 2 m from the pole, so h/(d + 2) = 1.5/2.
   2. [1 mark] Second position: the boy is d + 3 m from the pole and the tip is d + 3 + 3 = d + 6 m away, so h/(d + 6) = 1.5/3.
   3. [1 mark] (b) From the equations, h = 0.75(d + 2) and h = 0.5(d + 6); equating, 0.75d + 1.5 = 0.5d + 3 ⇒ 0.25d = 1.5.
   4. [1 mark] d = 6 m and h = 0.75 × 8 = 6 m (check: 0.5 × 12 = 6). The lamp is 6 m high.
   5. [1 mark] (c) Let the shadow be s m: 6/(15 + s) = 1.5/s ⇒ 6s = 22.5 + 1.5s ⇒ 4.5s = 22.5 ⇒ s = 5 m.

   Answer: d = 6 m; height of lamp = 6 m; shadow at 15 m = 5 m

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