CBSE Class 10 Mathematics Triangles MCQs Set 17

Answer: (b) TX² = TB × TC

Explanation:
1. In △TXM, BN ∥ XM, so \( \frac{TB}{TX}=\frac{TN}{TM} \).
2. In △TMC, XN ∥ CM, so \( \frac{TX}{TC}=\frac{TN}{TM} \).
3. Equating, \( \frac{TB}{TX}=\frac{TX}{TC} \), hence TX² = TB × TC.

Teacher's Note:
1. Apply the basic proportionality theorem twice in two triangles that share the ratio TN/TM.
2. Eliminating the common ratio gives a product relation.

Answer: (a) AD = BE

Explanation:
1. ∠A = ∠B, so the sides opposite them are equal: CB = CA.
2. \( DE\parallel AB\Rightarrow\frac{CD}{DA}=\frac{CE}{EB} \). Adding 1 to both sides: \( \frac{CA}{DA}=\frac{CB}{EB} \).
3. Since CA = CB, DA = EB.

Teacher's Note:
1. Equal angles give equal opposite sides.
2. Add 1 to both sides of the proportion to turn parts into whole sides.

Answer: (a) AD² = AB × AF

Explanation:
1. From DE ∥ BC in △ABC: \( \frac{AB}{AD}=\frac{AC}{AE} \).
2. From EF ∥ CD in △ADC: \( \frac{AD}{AF}=\frac{AC}{AE} \).
3. So \( \frac{AB}{AD}=\frac{AD}{AF} \), i.e. AD² = AB × AF.

Teacher's Note:
1. Use the basic proportionality theorem in two different triangles that share the ratio AC/AE.
2. The middle term in the proportion is squared.

Answer: (a) \( \frac1z=\frac1x+\frac1y \)

Explanation:
1. Let the feet of the poles be A and C, and let the vertical through Q meet AC at B, with AB = a and BC = b.
2. Triangles QBC and PAC are similar: \( \frac zx=\frac b{a+b} \). Triangles QBA and RCA are similar: \( \frac zy=\frac a{a+b} \).
3. Adding, \( \frac zx+\frac zy=1\Rightarrow\frac1x+\frac1y=\frac1z \).

Teacher's Note:
1. Drop the perpendicular from the crossing point and use similar right triangles on each side.
2. Check with x = 6, y = 3: z = 2 and 1/6 + 1/3 = 1/2.

Answer: (d) isosceles with AB = AC

Explanation:
1. AD/DB = AE/EC gives DE ∥ BC by the converse of the basic proportionality theorem.
2. Then ∠ADE = ∠ABC (corresponding angles).
3. Given ∠ADE = ∠ACB, so ∠ABC = ∠ACB and hence AB = AC.

Teacher's Note:
1. Converse of the basic proportionality theorem first, then corresponding angles.
2. Equal angles mean equal opposite sides.

Answer: (c) BC ∥ QR

Explanation:
1. In △OPQ, AB ∥ PQ, so OA/AP = OB/BQ.
2. In △OPR, AC ∥ PR, so OA/AP = OC/CR.
3. Hence OB/BQ = OC/CR, and by the converse of the basic proportionality theorem BC ∥ QR.

Teacher's Note:
1. Link the two ratios through OA/AP.
2. Equal ratios on two sides of △OQR give a parallel third side.

Answer: (b) EF ∥ BC

Explanation:
1. In △ADB, DE bisects ∠ADB, so AE/EB = AD/DB.
2. In △ADC, DF bisects ∠ADC, so AF/FC = AD/DC.
3. Since BD = DC, AE/EB = AF/FC, so EF ∥ BC (converse of the basic proportionality theorem).

Teacher's Note:
1. The median gives BD = DC, which makes the two bisector ratios equal.
2. Equal ratios on two sides mean the third side is parallel.

Answer: (a) AD × BE × CF = DB × EC × FA

Explanation:
1. OD bisects ∠AOB in △AOB: AD/DB = OA/OB.
2. OE bisects ∠BOC: BE/EC = OB/OC. OF bisects ∠COA: CF/FA = OC/OA.
3. Multiplying: \( \frac{AD}{DB}\cdot\frac{BE}{EC}\cdot\frac{CF}{FA}=\frac{OA}{OB}\cdot\frac{OB}{OC}\cdot\frac{OC}{OA}=1 \).

Teacher's Note:
1. Use the angle bisector theorem in the three small triangles at O.
2. The product of the three ratios cancels completely.

Answer: (d) AE = AC

Explanation:
1. Since CE ∥ DA: ∠DAC = ∠ACE (alternate angles) and ∠BAD = ∠AEC (corresponding angles).
2. AD bisects ∠A, so ∠BAD = ∠DAC. Therefore ∠ACE = ∠AEC.
3. In △ACE, equal angles give equal opposite sides: AE = AC. Then BD/DC = BA/AE = AB/AC.

Teacher's Note:
1. The construction turns the bisected angles into the base angles of an isosceles triangle.
2. This is how BD/DC = AB/AC is obtained.

Answer: (c) 60°

Explanation:
1. AB/LM = 4.4/11 = 2/5, BC/MN = 4/10 = 2/5, CA/NL = 3.6/9 = 2/5.
2. By SSS similarity △ABC ~ △LMN, so ∠L = ∠A.
3. ∠A = 180° − 50° − 70° = 60°.

Teacher's Note:
1. Match the sides in order A-L, B-M, C-N before comparing angles.
2. Equal side ratios mean equal corresponding angles.

Answer: (b) △DEF ~ △MNP by AA similarity

Explanation:
1. In △MNP, ∠N = 180° − 70° − 30° = 80°.
2. So ∠D = ∠M = 70° and ∠E = ∠N = 80°.
3. By AA similarity △DEF ~ △MNP, with the vertices in that order.

Teacher's Note:
1. Find the third angle first.
2. In a similarity statement, the vertex order shows which angles correspond.

Answer: (a) 15 cm

Explanation:
1. In △AOQ and △BOP, ∠OAQ = ∠OBP = 90° and ∠AOQ = ∠BOP (vertically opposite).
2. By AA similarity △AOQ ~ △BOP, so AO/BO = AQ/BP.
3. \( AQ=\frac{10\times9}{6}=15\ \text{cm} \).

Teacher's Note:
1. Vertically opposite angles plus right angles give AA similarity.
2. Write the ratio with the sides of the same triangle in the same order.

Answer: (d) 5.6 cm and 3.25 cm

Explanation:
1. From the similarity, AC/AP = CB/PQ = AB/AQ.
2. CB/PQ = 8/4 = 2, so AC = 2 × 2.8 = 5.6 cm.
3. AB/AQ = 2, so AQ = 6.5/2 = 3.25 cm.

Teacher's Note:
1. Read corresponding vertices from the order in the similarity statement.
2. The common ratio here is 2.

Answer: (c) 60 cm

Explanation:
1. Check the data: 100² + 75² = 125², so BC = 125 cm is consistent with a right angle at A.
2. Area of △ABC in two ways: ½ × AB × AC = ½ × BC × AD, so AD × BC = AB × AC.
3. \( AD=\frac{100\times75}{125}=60\ \text{cm} \).

Teacher's Note:
1. The altitude to the hypotenuse equals (product of the legs) / hypotenuse.
2. Check that the data are consistent before using them.

Answer: (b) 10 cm

Explanation:
1. ∠AOB = ∠COD (vertically opposite) and AO/OC = BO/OD.
2. By SAS similarity △AOB ~ △COD, so AB/DC = AO/OC = 1/2.
3. DC = 2 × 5 = 10 cm.

Teacher's Note:
1. SAS similarity needs one equal angle between two proportional sides.
2. The common ratio of sides is 1 : 2.

Answer: (a) CA² = CB × CD

Explanation:
1. In △ABC and △DAC, ∠BAC = ∠ADC (given) and ∠C is common.
2. So △ABC ~ △DAC (AA), giving CB/CA = CA/CD.
3. Hence CA² = CB × CD.

Teacher's Note:
1. Write the similarity so that the equal angles correspond (A with D, C with C).
2. Cross-multiply the proportion CB/CA = CA/CD.

Answer: (d) 3 : 1

Explanation:
1. AB = 9 cm, AC = 15 cm. AP/AB = 3/9 = 1/3 and AQ/AC = 5/15 = 1/3.
2. With the common angle A, △APQ ~ △ABC (SAS).
3. So PQ/BC = 1/3, i.e. BC = 3PQ.

Teacher's Note:
1. Add the parts to get the whole sides first.
2. Equal ratios on two sides mean PQ ∥ BC as well.

Answer: (c) 6 cm

Explanation:
1. In △CAB and △CED, ∠A = ∠CED and ∠C is common, so △CAB ~ △CED.
2. CA/CE = AB/DE = CB/CD, i.e. 15/10 = 9/x = 12/8.
3. x = 9 × 10/15 = 6 cm.

Teacher's Note:
1. Add the segments to get CA = 7 + 8 = 15 and CB = 10 + 2 = 12.
2. All three ratios equal 1.5, which confirms the answer.

Answer: (b) △ABD ~ △ECF by AA

Explanation:
1. AB = AC, so ∠B = ∠C.
2. ∠ADB = ∠EFC = 90°.
3. Hence △ABD ~ △ECF by AA similarity (A↔E, B↔C, D↔F).

Teacher's Note:
1. Isosceles triangle gives equal base angles.
2. A right angle in each triangle plus one more equal angle is enough for AA.

Answer: (a) AD² = BD × DC

Explanation:
1. In △ABD and △CAD, ∠ADB = ∠ADC = 90° and ∠DBA = ∠DAC (each is 90° − ∠BAD).
2. So △DBA ~ △DAC, giving DB/DA = DA/DC.
3. Hence AD² = BD × DC.

Teacher's Note:
1. The altitude to the hypotenuse splits the triangle into two triangles similar to each other.
2. AD is the mean proportional between BD and DC.

Answer: (d) DF × EF = FB × FA

Explanation:
1. In △AFD and △BFE, ∠AFD = ∠BFE (vertically opposite) and ∠FDA = ∠FBE (alternate angles, AD ∥ BC).
2. So △AFD ~ △BFE (F↔F, D↔B, A↔E), giving FD/FB = FA/FE.
3. Cross-multiplying, DF × EF = FB × FA.

Teacher's Note:
1. AD ∥ BC supplies the alternate angles.
2. Match vertices carefully: D with B and A with E.

Answer: (c) 2 BL

Explanation:
1. △BMC ≅ △EMD (AAS), so DE = BC. Also AD = BC, so AE = AD + DE = 2BC.
2. In △AEL and △CBL, ∠ALE = ∠CLB (vertically opposite) and ∠EAL = ∠BCL (alternate angles), so they are similar.
3. EL/BL = AE/CB = 2BC/BC = 2, so EL = 2BL.

Teacher's Note:
1. First prove DE = BC using the mid-point.
2. Then compare AE with CB through the second similarity.

Answer: (b) AD = BC

Explanation:
1. AB ∥ DC gives △EDC ~ △EBA, so ED/EB = EC/EA.
2. From △AED ~ △BEC: ED/EC = EA/EB = AD/BC.
3. Combining gives EA/EB = EB/EA, so EA = EB.
4. Then AD/BC = EA/EB = 1, so AD = BC.

Teacher's Note:
1. Use both similarities of the trapezium diagonals.
2. A trapezium with equal non-parallel sides is isosceles.

Answer: (a) 50 m

Explanation:
1. At the same time the sun's rays make the same angle, so the stick and tower with their shadows form similar right triangles.
2. h/15 = 20/6, so h = 15 × 20/6 = 50 m.

Teacher's Note:
1. Height : shadow is the same ratio for both objects.
2. Units cancel in the ratio, so cm and m need not be converted.

Answer: (d) 2√13 cm

Explanation:
1. △ADB ~ △ABC, so AB² = AD × AC.
2. AC = 4 + 9 = 13 cm, so AB² = 4 × 13 = 52.
3. AB = √52 = 2√13 cm.

Teacher's Note:
1. Each leg is the mean proportional between the hypotenuse and its projection on it.
2. BD = 6 cm is the geometric mean of AD and DC, not AB.