CBSE Class 10 Mathematics Triangles MCQs Set 11

Answer: (a) ΔAOB ~ ΔDOC because \( \frac{AO}{DO} = \frac{BO}{CO} \) and ∠AOB = ∠DOC

Explanation:
1. \( \frac{AO}{DO} = \frac{1.6}{0.8} = 2 \) and \( \frac{BO}{CO} = \frac{4.8}{2.4} = 2 \), so the sides including the angle at O are proportional.
2. ∠AOB = ∠DOC (vertically opposite angles).
3. So ΔAOB ~ ΔDOC by SAS similarity. A ratio alone, or one angle alone, is not enough.

Teacher's Note:
1. SAS similarity needs the angle between the two proportional sides.
2. Vertically opposite angles at the crossing point are always equal.

Answer: (b) BC · DE = AB · EF

Explanation:
1. ΔABC ~ ΔEDF gives \( \frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} \).
2. Cross-multiplying: BC · EF = AC · DF, AB · EF = AC · ED and BC · ED = AB · DF are all true.
3. BC · DE = AB · EF would need BC/AB = EF/DE, but the true ratio is BC/AB = DF/DE. So option (c) is not true.

Teacher's Note:
1. Match the vertices in order: A-E, B-D, C-F.
2. Write the three equal ratios first and then cross-multiply.

Answer: (c) XY = \( \frac{1}{3} \) QR

Explanation:
1. PX : XQ = 1 : 2, so PX : PQ = 1 : 3.
2. XY ∥ QR gives ΔPXY ~ ΔPQR (AA).
3. \( \frac{XY}{QR} = \frac{PX}{PQ} = \frac{1}{3} \), so XY = \( \frac{1}{3} \) QR.

Teacher's Note:
1. Convert the ratio PX : XQ into a ratio with the whole side PQ.
2. The ratio of corresponding sides of similar triangles equals the ratio of the parts from the vertex.

Answer: (d) 14 cm

Explanation:
1. ΔOPQ ~ ΔORS (PQ ∥ SR, alternate angles, and vertically opposite angles).
2. \( \frac{PQ}{SR} = \frac{PO}{OR} = \frac{1}{2} \).
3. SR = 2 × 7 = 14 cm.

Teacher's Note:
1. The diagonals of a trapezium cut each other in the ratio of the parallel sides.
2. Parallel sides PQ and SR are in the ratio 1 : 2.

Answer: (b) 4.05 cm

Explanation:
1. By the angle bisector theorem, \( \frac{QS}{SR} = \frac{PQ}{PR} = \frac{18}{22} \).
2. QS = \( 9 \times \frac{18}{18+22} = 9 \times \frac{9}{20} \) = 4.05 cm.

Teacher's Note:
1. The bisector divides the opposite side in the ratio of the other two sides.
2. Check: SR = 4.95 cm and 4.05 : 4.95 = 18 : 22.

Answer: (a) BD · CD = AD²

Explanation:
1. ΔADB ~ ΔCDA (both have a right angle and ∠BAD = ∠ACD).
2. \( \frac{BD}{AD} = \frac{AD}{CD} \), so AD² = BD · CD.

Teacher's Note:
1. The altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse.
2. Try a 3-4-5 triangle to check.

Answer: (c) 100°

Explanation:
1. \( \frac{PA}{PD} = \frac{6}{5} \) and \( \frac{PB}{PC} = \frac{3}{2.5} = \frac{6}{5} \), and ∠APB = ∠DPC.
2. So ΔAPB ~ ΔDPC (SAS) and ∠PBA = ∠PCD.
3. In ΔDPC: ∠PCD = 180° − 50° − 30° = 100°. Hence ∠PBA = 100°.

Teacher's Note:
1. Look for proportional sides around the vertically opposite angles.
2. Match B with C and A with D.

Answer: (d) 10 cm

Explanation:
1. The diagonals of a rhombus bisect each other at right angles, giving half-diagonals 8 cm and 6 cm.
2. Side = \( \sqrt{8^2+6^2} = \sqrt{100} \) = 10 cm.

Teacher's Note:
1. Each side is the hypotenuse of a right triangle made by two half-diagonals.
2. 8, 6, 10 is a 3-4-5 triple scaled by 2.

Answer: (a) ΔPQR ~ ΔCAB

Explanation:
1. AB ↔ QR gives A ↔ Q and B ↔ R; BC ↔ PR gives B ↔ R and C ↔ P; CA ↔ PQ gives C ↔ P and A ↔ Q.
2. So the correspondence is P ↔ C, Q ↔ A, R ↔ B, that is ΔPQR ~ ΔCAB (SSS).

Teacher's Note:
1. Read the matching of vertices from the ratios, not from the alphabetical order.
2. Each pair of equal ratios fixes one vertex.

Answer: (d) 3

Explanation:
1. ΔADE ~ ΔABC, so \( \frac{DE}{BC} = \frac{AD}{AB} = \frac{2}{5} \).
2. DE = \( 7.5 \times \frac{2}{5} \) = 3 cm.

Teacher's Note:
1. AB = AD + BD = 5 cm, not 3 cm.
2. Use AD/AB, not AD/BD.

Answer: (c) \( \frac{111}{7} \) cm

Explanation:
1. The lower ends lie on a straight line, so by similar triangles the rod length falls by the same amount for every 2 cm along the frame.
2. From hole 1 to hole 8 there are 7 gaps and the length falls from 29 cm to 6 cm, a fall of 23 cm. So each gap reduces the length by \( \frac{23}{7} \) cm.
3. Rod P is at hole 5, 4 gaps from hole 1: length = \( 29 - 4 \times \frac{23}{7} = \frac{203 - 92}{7} = \frac{111}{7} \) cm.

Teacher's Note:
1. Lengths of parallel rods at equal spacing form an A.P.
2. The line through the ends makes similar triangles with the frame.

Answer: (b) \( \frac{16}{3} \) cm

Explanation:
1. Let the side of the square be s. Then FD ∥ BE, so ΔAFD ~ ΔABE.
2. \( \frac{AF}{AB} = \frac{FD}{BE} \), that is \( \frac{16-s}{16} = \frac{s}{8} \).
3. 16 − s = 2s, so s = \( \frac{16}{3} \) cm.

Teacher's Note:
1. AF = AB − FB = 16 − s.
2. Check: 16 − 16/3 = 32/3 and 32/3 ÷ 16 = 2/3 = (16/3) ÷ 8.

Answer: (c) \( \frac{77}{117} \)

Explanation:
1. In right ΔADB: AB = \( \sqrt{33^2+56^2} = \sqrt{4225} \) = 65 cm.
2. In right ΔBAC: AC = \( \sqrt{97^2-65^2} = \sqrt{5184} \) = 72 cm.
3. Perimeter of ΔADB = 33 + 56 + 65 = 154 cm and perimeter of ΔABC = 65 + 72 + 97 = 234 cm.
4. Ratio = \( \frac{154}{234} = \frac{77}{117} \).

Teacher's Note:
1. The two triangles are not similar, so the perimeters must be found separately.
2. Both are right triangles, so use the Pythagoras theorem.

Answer: (d) The ratio of their altitudes is 4 : 7

Explanation:
1. For similar triangles, the ratios of perimeters, corresponding altitudes, medians and angle bisectors all equal the ratio of corresponding sides, 4 : 7.
2. So the ratio of altitudes is 4 : 7, and the other three statements are false.

Teacher's Note:
1. Every linear measurement is in the same ratio as the sides.
2. Only areas are in the ratio of squares, 16 : 49.

Answer: (a) \( \frac{EF}{RP} = \frac{DE}{PQ} \)

Explanation:
1. ∠D = ∠Q and ∠E = ∠R, so ∠F = ∠P and ΔDEF ~ ΔQRP (D ↔ Q, E ↔ R, F ↔ P).
2. Corresponding sides: DE ↔ QR, EF ↔ RP, DF ↔ QP, so \( \frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{QP} \).
3. Options (a), (c), (d) follow from this. In option (b), DE is divided by PQ instead of QR, so it is not true.

Teacher's Note:
1. Write the vertex matching D-Q, E-R, F-P before forming ratios.
2. The side opposite each matched vertex also matches.

Answer: (b) No, ∠D = ∠R is true but ∠F = ∠P is not true

Explanation:
1. ΔDEF ~ ΔRPQ means D ↔ R, E ↔ P and F ↔ Q.
2. So ∠D = ∠R is true, but ∠F = ∠Q (not ∠P). Hence ∠F = ∠P is not true in general.

Teacher's Note:
1. The order of letters tells which angles are equal.
2. Third letter F pairs with third letter Q.

Answer: (b) False, their corresponding sides must also be proportional

Explanation:
1. Similar polygons need equal corresponding angles and proportional corresponding sides.
2. For triangles, equal angles force proportional sides, but for quadrilaterals they do not: a square and a non-square rectangle have equal angles but are not similar.

Teacher's Note:
1. A rectangle 2 × 1 and a square both have four right angles.
2. Both conditions are needed for polygons with more than three sides.

Answer: (d) Yes, by AA similarity (∠D = ∠C and ∠A is common)

Explanation:
1. In ΔADE and ΔACB: ∠ADE = ∠ACB (given) and ∠DAE = ∠CAB (common angle).
2. So ΔADE ~ ΔACB by AA similarity. SSS cannot be used because no sides are given.

Teacher's Note:
1. Spot the common angle at A first.
2. Two equal angle pairs are enough for similarity.

Answer: (a) Yes, because \( \frac{PE}{QE} = \frac{PF}{RF} \)

Explanation:
1. \( \frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9} \) and \( \frac{PF}{RF} = \frac{8}{9} \).
2. The line EF divides PQ and PR in the same ratio, so by the converse of the basic proportionality theorem EF ∥ QR.

Teacher's Note:
1. Compare the ratio of the two parts on each side.
2. Do not compare PE with PF directly.

Answer: (c) Yes, by SAS similarity

Explanation:
1. \( \frac{PB}{PD} = \frac{5}{10} = \frac{1}{2} \) and \( \frac{PC}{PE} = \frac{6}{12} = \frac{1}{2} \).
2. ∠BPC = ∠DPE (vertically opposite angles).
3. So ΔPBC ~ ΔPDE by SAS similarity.

Teacher's Note:
1. Two ratios equal and the angle between them equal: SAS.
2. Only two sides of each triangle are given, so SSS cannot be used.

Answer: (d) Yes, the third sides are also in the ratio 3 : 1 (SSS similarity)

Explanation:
1. Let the sides of the second triangle be a, b, c and the corresponding sides of the first be 3a, 3b, y.
2. Perimeter: 3a + 3b + y = 3(a + b + c), so y = 3c.
3. All three sides are in the same ratio, so the triangles are similar by SSS.

Teacher's Note:
1. Use the perimeter to find the third side.
2. Congruent triangles would have a ratio 1 : 1, not 3 : 1.

Answer: (b) No, because \( \frac{ER}{RD} \neq \frac{ES}{SF} \)

Explanation:
1. \( \frac{ER}{RD} = \frac{5}{2.5} = 2 \) and \( \frac{ES}{SF} = \frac{1.5}{3.5} = \frac{3}{7} \).
2. The ratios are not equal, so by the converse of the basic proportionality theorem RS is not parallel to DF.

Teacher's Note:
1. RS ∥ DF only if both sides are cut in the same ratio.
2. Check by cross-multiplying: 5 × 3.5 ≠ 2.5 × 1.5.

Answer: (c) False, the equal angle must be the angle between the two proportional sides

Explanation:
1. SAS similarity needs the equal angle to be included between the two proportional sides.
2. If the angle is not between those sides, the triangles need not be similar. So the statement is not correct.

Teacher's Note:
1. The word “included” matters in SAS.
2. Try drawing two triangles with the same non-included angle and proportional sides.

Answer: (a) 10 m

Explanation:
1. At the same time the sun's rays make the same angle with the ground, so the tower and the pole with their shadows form similar triangles.
2. \( \frac{15}{x} = \frac{24}{16} \), so x = \( \frac{15 \times 16}{24} \) = 10 m.

Teacher's Note:
1. Height : shadow is the same ratio for both objects.
2. Check: 15/24 = 10/16 = 5/8.

Answer: (b) 8 cm, 12 cm, 16 cm

Explanation:
1. Produce SP to meet l at E. Since PA ∥ QB ∥ RC ∥ SD, they cut the transversals l and ES in the same ratio: PQ : QR : RS = AB : BC : CD = 6 : 9 : 12.
2. Let PQ = 6x, QR = 9x, RS = 12x. Then 27x = 36 and x = \( \frac{4}{3} \).
3. PQ = 8 cm, QR = 12 cm, RS = 16 cm.

Teacher's Note:
1. Parallel lines cut every transversal in the same ratio.
2. The three parts must add to 36 cm.

Answer: (a) 4

Explanation:
1. By the basic proportionality theorem, \( \frac{AD}{DB} = \frac{AE}{EC} \).
2. \( \frac{x}{x-2} = \frac{x+2}{x-1} \) gives x(x − 1) = (x + 2)(x − 2), so x² − x = x² − 4.
3. Hence x = 4.

Teacher's Note:
1. Cross-multiply and cancel x² from both sides.
2. Check: 4/2 = 6/3 = 2.

Answer: (c) 120 m

Explanation:
1. The pole and the tree with their shadows form similar right triangles.
2. \( \frac{24}{16} = \frac{h}{80} \), so h = \( \frac{24 \times 80}{16} \) = 120 m.

Teacher's Note:
1. The ratio height : shadow is 3 : 2 for both.
2. 80 × 3/2 = 120.

Answer: (d) 170 m

Explanation:
1. P, O, S are in a line and Q, O, R are in a line, so ∠POQ = ∠SOR (vertically opposite).
2. \( \frac{PO}{SO} = \frac{180}{90} = 2 \) and \( \frac{QO}{RO} = \frac{140}{70} = 2 \), so ΔPOQ ~ ΔSOR (SAS).
3. \( \frac{PQ}{SR} = 2 \), so PQ = 2 × 85 = 170 m.

Teacher's Note:
1. Equal ratios 2 and 2 on both sides of the vertical angle.
2. PQ corresponds to SR.

Answer: (b) \( \frac{xz}{y+z} \)

Explanation:
1. ∠LMK = ∠PNK = 55° (corresponding angles), so PN ∥ LM.
2. ΔKNP ~ ΔKML (AA, common ∠K). \( \frac{NP}{ML} = \frac{NK}{MK} \).
3. \( \frac{a}{x} = \frac{z}{y+z} \), so a = \( \frac{xz}{y+z} \).

Teacher's Note:
1. MK = MN + NK = y + z.
2. Equal corresponding angles mean parallel lines.

Answer: (d) \( \frac{3}{4} \)

Explanation:
1. QG ∥ PR gives ∠GQF = ∠PRE (alternate angles), and BF ∥ PE gives ∠QFG = ∠REP.
2. So ΔQFG ~ ΔREP (AA).
3. \( \frac{FG}{EP} = \frac{QG}{RP} = \frac{3}{4} \).

Teacher's Note:
1. Name the triangles with the vertices in matching order: Q-R, F-E, G-P.
2. The ratio of similarity equals QG : PR.

Answer: (c) \( \frac{9}{4} \) cm

Explanation:
1. From ΔQFG ~ ΔREP: \( \frac{FQ}{ER} = \frac{FG}{EP} \).
2. \( \frac{FG}{EP} = \frac{2}{8/3} = \frac{3}{4} \), so FQ = \( \frac{3}{4} \times 3 = \frac{9}{4} \) cm.

Teacher's Note:
1. FQ matches ER in the two similar triangles.
2. This agrees with the ratio 3/4 found from QG/PR.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In ΔAEC and ΔBED: ∠ACE = ∠BDE = 90° and ∠AEC = ∠BED (vertically opposite). So ΔAEC ~ ΔBED by AA.
2. \( \frac{AE}{BE} = \frac{CE}{DE} \), so AE × ED = CE × EB. Assertion is true.
3. Reason is the AA criterion used here, so it explains the Assertion.

Teacher's Note:
1. Equal angles give similar triangles, and similar triangles give proportional sides.
2. Cross-multiply the ratio to get the product form.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Let the sides of the second triangle be a, b, c and two sides of the first be 3a, 3b. If the third side is y, then 3a + 3b + y = 3(a + b + c), so y = 3c.
2. All three sides are in the ratio 3 : 1, so the triangles are similar by SSS. Assertion is true.
3. Reason is the SSS criterion, which explains the Assertion.

Teacher's Note:
1. The perimeter condition forces the third side to be 3 times as well.
2. SSS similarity needs all three ratios equal.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. PQ is a radius, so PQ = 6 cm. AE ∥ BQ gives ΔPAE ~ ΔPBQ (AA).
2. \( \frac{PE}{PQ} = \frac{PA}{PB} = \frac{3}{4} \), so PE = 4.5 cm. Assertion is true.
3. Reason is the AA criterion used to get the similarity, so it explains the Assertion.

Teacher's Note:
1. A radius of the sector gives PQ = PR = 6 cm.
2. The common angle at P and the parallel lines give two equal angle pairs.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Through A draw AD ∥ QC meeting PR at D. A is the mid-point of QR, so D is the mid-point of CR, that is CD = DR.
2. In ΔPAD, B is the mid-point of PA and BC ∥ AD, so C is the mid-point of PD, that is PC = CD.
3. So PC = CD = DR and PC = \( \frac{1}{3} \) PR. Assertion is true.
4. The Reason is also a true statement, but it is not used in the proof, so it does not explain the Assertion.

Teacher's Note:
1. Use the mid-point theorem twice.
2. Both statements true but unrelated: choose (b).