CBSE Class 10 Mathematics Triangles MCQs Set 13

Answer: (c) 5 : 6

Explanation:
1. The triangles are similar, so the ratio of areas is the square of the ratio of corresponding altitudes.
2. \( \frac{h_1^2}{h_2^2}=\frac{25}{36} \), so \( \frac{h_1}{h_2}=\frac{5}{6} \).

Teacher's Note:
1. Area ratio = (side ratio)²; the same holds for heights.
2. Take the square root of 25 : 36.

Answer: (a) 4.5 cm

Explanation:
1. By the angle bisector theorem, \( \frac{AB}{AC}=\frac{BD}{DC} \).
2. \( \frac{6}{AC}=\frac{4}{3} \), so AC = 4.5 cm.

Teacher's Note:
1. The bisector divides the opposite side in the ratio of the other two sides.
2. Cross-multiply carefully.

Answer: (c) 36\( \sqrt{2} \) cm

Explanation:
1. Let AB = x. Then BF = x/2 and BE = x/3, so the area of △FBE = \( \frac{1}{2}\cdot\frac{x}{2}\cdot\frac{x}{3}=\frac{x^2}{12} \).
2. x²/12 = 108 gives x = 36 cm.
3. The diagonal AC = \( 36\sqrt{2} \) cm.

Teacher's Note:
1. The diagonal of a square is side × √2.
2. Use the right angle at B for the triangle area.

Answer: (a) 21 m

Explanation:
1. On one side: AC² = 15² − 9² = 144, so AC = 12 m.
2. On the other side: CB² = 15² − 12² = 81, so CB = 9 m.
3. Width = 12 + 9 = 21 m.

Teacher's Note:
1. Use Pythagoras in each right triangle.
2. Add the two horizontal distances.

Answer: (a) All isosceles triangles are similar.

Explanation:
1. Equilateral triangles have all angles 60°, so they are similar; all circles are similar.
2. Isosceles triangles can have different angles, so they need not be similar.

Teacher's Note:
1. Similar means equal angles and proportional sides.
2. Look for a counter-example.

Answer: (b) 18 cm

Explanation:
1. Let the legs be a, b and the hypotenuse c: ab = 80, a + b + c = 40, a² + b² = c².
2. (a + b)² = (40 − c)² = c² + 160, so 1600 − 80c = 160 and c = 18.
3. Check: a + b = 22 and ab = 80 give a² + b² = 484 − 160 = 324 = 18².

Teacher's Note:
1. Use (a + b)² = a² + b² + 2ab.
2. Eliminate a and b together.

Answer: (a) 14 cm²

Explanation:
1. Ratio of areas = (AB/DE)² = 4.
2. Area of △DEF = 56 ÷ 4 = 14 cm².

Teacher's Note:
1. Area ratio is the square of the side ratio.
2. Divide the larger area by 4.

Answer: (c) 9 cm

Explanation:
1. Scale factor = A′B′/AB = 15/5 = 3.
2. A′C′ = 3 × AC = 9 cm.

Teacher's Note:
1. Find the scale factor from AB and A′B′.
2. Multiply a side of △ABC by it.

Answer: (a) 4 cm

Explanation:
1. BC = B′C′ ÷ 3 = 12 ÷ 3 = 4 cm.

Teacher's Note:
1. Divide by the scale factor.
2. Corresponding sides are in the same ratio.

Answer: (b) 80°

Explanation:
1. Similar triangles have equal corresponding angles, so ∠A′ = ∠A = 80°.

Teacher's Note:
1. Corresponding angles are equal.
2. No scaling is needed for angles.

Answer: (a) 60°

Explanation:
1. ∠B = ∠B′ = 60°.

Teacher's Note:
1. Use equal corresponding angles.
2. B corresponds to B′.

Answer: (b) 40°

Explanation:
1. ∠C = 180° − 80° − 60° = 40°.

Teacher's Note:
1. Use the angle sum property.
2. ∠C = ∠C′ as well.

Answer: (b) 24 cm, 9 cm

Explanation:
1. Side ratio = √(16/9) = 4/3.
2. AB = 18 × 4/3 = 24 cm and QR = 12 × 3/4 = 9 cm.

Teacher's Note:
1. Take the square root of the area ratio.
2. AB corresponds to PQ and BC corresponds to QR.

Answer: (d) 3 : 20

Explanation:
1. BE = AB/4 and BF = (3/5)BC.
2. Triangles BEF and BAC share angle B, so the area ratio = (BE × BF)/(BA × BC) = 1/4 × 3/5 = 3/20.

Teacher's Note:
1. Triangles with a common angle have areas in the ratio of the products of the sides including it.
2. Check which segment is the shorter part.

Answer: (d) 15 cm

Explanation:
1. For similar triangles, the ratio of perimeters equals the ratio of corresponding sides.
2. AB/PQ = 36/24 = 3/2, so AB = 15 cm.

Teacher's Note:
1. Perimeters are in the same ratio as sides.
2. AB corresponds to PQ.

Answer: (a) 9

Explanation:
1. Area ratio = (QR/BC)² = 3² = 9.

Teacher's Note:
1. Area ratio is the square of the side ratio.
2. Notice that PQR is the larger triangle.

Answer: (c) 9 : 7

Explanation:
1. Ratio of areas = (ratio of medians)² = 81 : 49.
2. Ratio of medians = 9 : 7.

Teacher's Note:
1. Take the square root of the area ratio.
2. Medians behave like sides.

Answer: (b) \( \sqrt{3}\,p \)

Explanation:
1. D is the mid-point of BC, so BD = p.
2. AD² = (2p)² − p² = 3p², so AD = √3 p.

Teacher's Note:
1. The altitude of an equilateral triangle bisects the base.
2. Use Pythagoras in △ABD.

Answer: (d) 6 cm

Explanation:
1. BC/QR = √(9/16) = 3/4.
2. QR = 4.5 × 4/3 = 6 cm.

Teacher's Note:
1. Square root of the area ratio gives the side ratio.
2. BC corresponds to QR.

Answer: (b) 1 : 2

Explanation:
1. ar(APQ)/ar(ABC) = 4, so PQ/BC = 2.
2. BC : PQ = 1 : 2.

Teacher's Note:
1. Area ratio is the square of the side ratio.
2. Note which triangle is larger.