CBSE Class 10 Mathematics Triangles MCQs Set 14

Answer: (a) ΔPQR ~ ΔCAB

Explanation:
1. AB ↔ QR, BC ↔ PR, CA ↔ PQ, so A ↔ Q, B ↔ R, C ↔ P.
2. Writing the vertices of ΔPQR in order: P ↔ C, Q ↔ A, R ↔ B, giving ΔPQR ~ ΔCAB.

Teacher's Note:
1. Match the vertices from the common letters of the sides.
2. The order of letters in a similarity statement must follow the correspondence.

Answer: (b) 81 : 25

Explanation:
1. Ratio of areas = (ratio of corresponding sides)².
2. (9 : 5)² = 81 : 25.

Teacher's Note:
1. Areas of similar figures go as the square of corresponding sides.
2. Keep the order of the triangles.

Answer: (d) 49 : 81

Explanation:
1. Ratio of sides = ratio of perimeters = 35 : 45 = 7 : 9.
2. Ratio of areas = 7² : 9² = 49 : 81.

Teacher's Note:
1. Perimeters of similar triangles are in the ratio of their sides.
2. Then square it for the areas.

Answer: (a) 5.4 cm

Explanation:
1. Ratio of corresponding sides = 25 : 15 = 5 : 3.
2. Required side = 9 × 3 ÷ 5 = 5.4 cm.

Teacher's Note:
1. Perimeter ratio equals side ratio.
2. The second triangle is the smaller one.

Answer: (b) 7.5 cm

Explanation:
1. PQ ↔ XY and QR ↔ YZ, so \( \frac{PQ}{XY} = \frac{QR}{YZ} \).
2. \( \frac{4}{6} = \frac{5}{YZ} \) gives YZ = 7.5 cm.

Teacher's Note:
1. Write the proportion with corresponding sides in the same order.
2. Cross-multiply to solve.

Answer: (b) 22 cm

Explanation:
1. \( \frac{AB^2}{DE^2} = \frac{169}{121} \), so \( \frac{AB}{DE} = \frac{13}{11} \).
2. DE = 26 × 11 ÷ 13 = 22 cm.

Teacher's Note:
1. Take the square root of the area ratio to get the side ratio.
2. 13² = 169 and 11² = 121.

Answer: (b) 9 : 49

Explanation:
1. PR = PS + SR = 7 cm and ΔPST ~ ΔPRQ.
2. Ratio of areas = PS² : PR² = 9 : 49.

Teacher's Note:
1. Use the whole side PR = 7, not SR = 4.
2. The small triangle has side PS.

Answer: (a) PB × DP

Explanation:
1. ΔAPB ~ ΔDPC (∠A = ∠D = 90°, vertically opposite angles at P).
2. \( \frac{AP}{DP} = \frac{PB}{PC} \), so AP × PC = PB × DP.

Teacher's Note:
1. Look for the pair of similar triangles first.
2. Cross-multiply the proportion.

Answer: (b) \( \frac{AD}{PM} \)

Explanation:
1. Corresponding medians of similar triangles are in the same ratio as corresponding sides.
2. So \( \frac{AB}{PQ} = \frac{AD}{PM} \).

Teacher's Note:
1. The ratio of medians equals the ratio of sides.
2. Pick the pair of corresponding medians.

Answer: (c) 21 cm²

Explanation:
1. ΔAOB ~ ΔCOD, with AB : CD = 2 : 1.
2. Areas are in the ratio 4 : 1, so ar(COD) = 84 ÷ 4 = 21 cm².

Teacher's Note:
1. Triangles formed by the diagonals of a trapezium are similar.
2. Square the ratio of the parallel sides.

Answer: (c) 96 cm²

Explanation:
1. \( \frac{54}{ar(DEF)} = \frac{3^2}{4^2} = \frac{9}{16} \).
2. ar(DEF) = 54 × 16 ÷ 9 = 96 cm².

Teacher's Note:
1. Areas are in the ratio of the squares of corresponding sides.
2. The larger sides give the larger area.

Answer: (c) 4.5 cm

Explanation:
1. Altitude = (√3/2) × side = (√3/2) × 3√3.
2. = 9/2 = 4.5 cm.

Teacher's Note:
1. Use altitude = (√3/2)a.
2. √3 × √3 = 3.

Answer: (c) 2AC²

Explanation:
1. AC = BC and ∠C = 90°, so AB² = AC² + BC² = AC² + AC².
2. AB² = 2AC².

Teacher's Note:
1. Use Pythagoras theorem with AC = BC.
2. The hypotenuse square is twice the square of a leg.

Answer: (c) 100√269 km

Explanation:
1. Distances covered: 1000 km north and 1300 km east, at right angles.
2. Distance = √(1000² + 1300²) = √2690000 = 100√269 km.

Teacher's Note:
1. Distance = speed × time for each plane.
2. Then apply Pythagoras theorem.

Answer: (c) 300√61 km

Explanation:
1. North: 1500 km, west: 1800 km.
2. Distance = √(1500² + 1800²) = 300√(25 + 36) = 300√61 km.

Teacher's Note:
1. Take 300 as a common factor.
2. North and west are perpendicular.

Answer: (c) 24 cm²

Explanation:
1. Area of a kite = ½ × product of diagonals.
2. = ½ × 6 × 8 = 24 cm².

Teacher's Note:
1. The sticks are the diagonals of the kite.
2. Perpendicular diagonals give area ½ d₁d₂.

Answer: (c) 15 cm

Explanation:
1. Scale factor model : boat = 1 : 4.
2. Model width = 60 ÷ 4 = 15 cm.

Teacher's Note:
1. A scale of 1 : 4 means the model is one quarter of the real size.
2. Apply it to the width, not to the length.

Answer: (d) 5 m

Explanation:
1. Height and shadow are in the same ratio at the same time (similar right triangles).
2. Shadow of tree = 12.5 × 2 ÷ 5 = 5 m.

Teacher's Note:
1. The sun makes the same angle with the ground for both.
2. Use the proportion height : shadow.

Answer: (d) 18 cm

Explanation:
1. By the basic proportionality theorem AD/DB = AE/EC = 1/3.
2. EC = 3 × 4.5 = 13.5 cm, so AC = 4.5 + 13.5 = 18 cm.

Teacher's Note:
1. Find EC first, then add AE.
2. AC = AE + EC.

Answer: (b) isosceles triangle

Explanation:
1. AD/DB = AE/EC gives DE ∥ BC (converse of the basic proportionality theorem).
2. So ∠B = ∠ADE = ∠AED = ∠C, hence AB = AC: isosceles.

Teacher's Note:
1. Equal angles mean equal opposite sides.
2. Corresponding angles with parallel lines are equal.