CBSE Class 10 Mathematics Triangles MCQs Set 15

Answer: (b) 15 cm

Explanation:
1. Since DE ∥ BC, the Basic Proportionality Theorem gives \( \frac{AD}{DB} = \frac{AE}{EC} \).
2. So \( \frac{AE}{EC} = \frac{2}{3} \). With AE = 6 cm, EC = 9 cm.
3. AC = AE + EC = 6 + 9 = 15 cm.

Teacher's Note:
1. Add the two parts to get the whole side.
2. The ratio AD : DB = AE : EC holds for a line parallel to BC.

Answer: (b) 15 cm

Explanation:
1. The diagonals of a rhombus bisect each other at right angles, so half-diagonals are 9 cm and 12 cm.
2. Side = \( \sqrt{9^2 + 12^2} = \sqrt{225} \) = 15 cm.

Teacher's Note:
1. The side is the hypotenuse of the right triangle with the half-diagonals as legs.
2. 9, 12, 15 is a multiple of the triple 3, 4, 5.

Answer: (c) 6 cm

Explanation:
1. AB = 4BX means AX = AB − BX = 3BX, so AX : XB = 3 : 1.
2. Since XY ∥ BC, \( \frac{AY}{YC} = \frac{AX}{XB} = 3 \).
3. With YC = 2 cm, AY = 3 × 2 = 6 cm.

Teacher's Note:
1. First find AX : XB from AB = 4BX.
2. Then apply the Basic Proportionality Theorem.

Answer: (b) 8 cm

Explanation:
1. Since AB ∥ DC, △AOB ∼ △COD (AA).
2. \( \frac{AB}{DC} = \frac{AO}{OC} = \frac{1}{2} \).
3. With AB = 4 cm, DC = 8 cm.

Teacher's Note:
1. Diagonals of a trapezium divide each other in the ratio of the parallel sides.
2. Vertically opposite angles at O are equal.

Answer: (b) 5AC²

Explanation:
1. In right triangle ABN, AN² = AB² + BN² = AB² + \( \frac{BC^2}{4} \).
2. In right triangle CBM, CM² = BC² + BM² = BC² + \( \frac{AB^2}{4} \).
3. AN² + CM² = \( \frac{5}{4} \)(AB² + BC²) = \( \frac{5}{4} \)AC².
4. So 4(AN² + CM²) = 5AC².

Teacher's Note:
1. Use the Pythagoras theorem twice.
2. AB² + BC² = AC² as the angle at B is a right angle.

Answer: (c) 8.8 cm

Explanation:
1. Since LM ∥ QR, \( \frac{PM}{MR} = \frac{PL}{LQ} = \frac{1}{3} \).
2. With MR = 6.6 cm, PM = 6.6 ÷ 3 = 2.2 cm.
3. PR = PM + MR = 2.2 + 6.6 = 8.8 cm.

Teacher's Note:
1. PM : MR = 1 : 3, so PM is one third of MR.
2. Add PM and MR for the whole side.

Answer: (b) 56°

Explanation:
1. For similar triangles ABC and DEF, the angles correspond in order: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.
2. So ∠C = ∠F = 56°.

Teacher's Note:
1. Corresponding angles of similar triangles are equal.
2. The order of letters in the similarity statement gives the correspondence.

Answer: (a) 20 cm

Explanation:
1. Half-diagonals are 12 cm and 16 cm and they meet at right angles.
2. Side = \( \sqrt{12^2 + 16^2} = \sqrt{400} \) = 20 cm.

Teacher's Note:
1. 12, 16, 20 is 4 times the triple 3, 4, 5.
2. The four sides of a rhombus are equal.

Answer: (b) △CBA ∼ △PQR

Explanation:
1. \( \frac{AB}{RQ} = \frac{BC}{QP} = \frac{CA}{PR} \) shows that A ↔ R, B ↔ Q and C ↔ P.
2. So △ABC ∼ △RQP. Writing the same correspondence in another order gives △CBA ∼ △PQR.

Teacher's Note:
1. In a similarity statement, letters in the same position correspond.
2. Check the other options: they put different vertices in the same place.

Answer: (c) 30 cm

Explanation:
1. CA = 2.5 cm corresponds to DF = 7.5 cm, so the scale factor is \( \frac{7.5}{2.5} \) = 3.
2. Perimeter of ABC = 4 + 3.5 + 2.5 = 10 cm.
3. Perimeter of DEF = 3 × 10 = 30 cm.

Teacher's Note:
1. Perimeters of similar triangles are in the same ratio as the corresponding sides.
2. CA corresponds to FD because ABC ∼ DEF.

Answer: (a) 100 m

Explanation:
1. The stick and its shadow, and the tower and its shadow, form similar right triangles (same time of the day).
2. \( \frac{\text{height of tower}}{50} = \frac{20}{10} \).
3. Height of the tower = 100 m.

Teacher's Note:
1. The sun's rays make the same angle with the ground at the same time.
2. Equal ratios of height to shadow.

Answer: (c) 10 cm

Explanation:
1. Since AB ∥ CD, △AOB ∼ △COD.
2. \( \frac{AB}{DC} = \frac{AO}{OC} = \frac{1}{2} \), so DC = 2 × AB = 10 cm.

Teacher's Note:
1. Use the similar triangles formed by the diagonals.
2. DC is the longer parallel side.

Answer: (c) similar but may not be congruent

Explanation:
1. Any two circles have the same shape, so they are similar.
2. They are congruent only if their radii are equal, which is not always the case.

Teacher's Note:
1. All circles are similar; all squares are similar.
2. Congruent means same shape and same size.

Answer: (c) \( \frac{PQ}{XZ} \) = \( \frac{QR}{YZ} \)

Explanation:
1. △PRQ ∼ △XYZ gives P ↔ X, R ↔ Y and Q ↔ Z.
2. So \( \frac{PR}{XY} = \frac{RQ}{YZ} = \frac{PQ}{XZ} \).
3. Option (c), \( \frac{PQ}{XZ} = \frac{QR}{YZ} \), is part of this chain.

Teacher's Note:
1. Match the sides by the vertices they join: PR with XY, RQ with YZ, PQ with XZ.
2. Check each option vertex by vertex.

Answer: (a) \( \frac{AB}{XY} \) = \( \frac{BC}{YZ} \)

Explanation:
1. ∠B = ∠X and ∠C = ∠Z, so ∠A = ∠Y, and △ABC ∼ △YXZ.
2. So \( \frac{AB}{YX} = \frac{BC}{XZ} = \frac{CA}{ZY} \).
3. In option (a), BC is divided by YZ, but BC corresponds to XZ, so (a) is not true in general. Options (b), (c) and (d) are parts of the chain.

Teacher's Note:
1. First find the third angles, then write the correspondence.
2. A wrong pairing of sides makes the statement false.

Answer: (b) similar but not congruent

Explanation:
1. In △LMN and △PQR, ∠L = ∠P and ∠N = ∠R, so the triangles are similar by AA.
2. MN corresponds to QR and MN = 2QR, so the scale factor is 2 and the sizes differ.
3. So they are similar but not congruent.

Teacher's Note:
1. Similar triangles are congruent only when the scale factor is 1.
2. MN and QR are the sides opposite the equal angles ∠L and ∠P.

Answer: (c) 50°

Explanation:
1. Compare the sides: \( \frac{AB}{QR} = \frac{4.5}{9} = \frac{1}{2} \), \( \frac{BC}{PQ} = \frac{5}{10} = \frac{1}{2} \), \( \frac{CA}{RP} = \frac{6\sqrt{2}}{12\sqrt{2}} = \frac{1}{2} \).
2. By SSS, △ABC ∼ △QRP, so C corresponds to P.
3. ∠C = 180° − 75° − 55° = 50°, so ∠P = 50°.

Teacher's Note:
1. Match the sides by size before naming the correspondence.
2. The vertex between the sides BC and CA is C, and the vertex between PQ and RP is P.

Answer: (a) \( \frac{4}{25} \)

Explanation:
1. \( \frac{ar(PQR)}{ar(XYZ)} = \left( \frac{PQ}{XY} \right)^2 = \frac{25}{4} \).
2. So \( \frac{ar(XYZ)}{ar(PQR)} = \frac{4}{25} \).

Teacher's Note:
1. Read which triangle is in the numerator.
2. Square the ratio of the sides, then invert if needed.

Answer: (c) a right angle

Explanation:
1. AB² = (6√7)² = 252 and CA² = 18² = 324, so AB² + CA² = 576.
2. BC² = 24² = 576, so AB² + CA² = BC².
3. By the converse of the Pythagoras theorem, ∠A = 90°.

Teacher's Note:
1. The longest side is opposite the right angle.
2. Compare the sum of the squares of the two shorter sides with the square of the longest.

Answer: (a) 24 cm

Explanation:
1. \( \frac{ar(DEF)}{ar(ABC)} = \frac{DF^2}{AC^2} = \frac{9}{16} \), so \( \frac{DF}{AC} = \frac{3}{4} \).
2. AC = 18 × \( \frac{4}{3} \) = 24 cm.

Teacher's Note:
1. DF and AC are corresponding sides because ABC ∼ DEF.
2. Take square roots of the area ratio.

Answer: (c) 25 cm

Explanation:
1. Half-diagonals are 15 cm and 20 cm, meeting at right angles.
2. Side = \( \sqrt{15^2 + 20^2} = \sqrt{625} \) = 25 cm.

Teacher's Note:
1. 15, 20, 25 is 5 times the triple 3, 4, 5.
2. All sides of a rhombus are equal.

Answer: (c) Two photographs of the same size of the same person, taken at the age of 20 years and at the age of 45 years, are not similar.

Explanation:
1. Similar figures need the same shape. A person's face changes between 20 and 45 years, so the two photographs do not have the same shape. (c) is true.
2. Congruent figures have the same shape and size, so they are always similar. (a) is false.
3. A circle and a square have different shapes. (b) is false.
4. A square and a rectangle have equal angles but their sides are not in proportion in general. (d) is false.

Teacher's Note:
1. Similar needs equal angles and proportional sides.
2. Photos of the same size only share the size, not the shape.

Answer: (d) All similar figures are congruent.

Explanation:
1. Similar figures have the same shape but may differ in size. Congruent figures need equal size as well.
2. For example, two circles of radii 2 cm and 3 cm are similar but not congruent. So (d) is false.
3. Statements (a), (b) and (c) are all true.

Teacher's Note:
1. Congruent implies similar, but similar does not imply congruent.
2. An enlarged photograph is similar to the original but not congruent.