CBSE Class 10 Mathematics Triangles MCQs Set 12

Answer: (a) 4.5 cm

Explanation:
1. AD bisects ∠A, so AB/AC = BD/DC (angle bisector theorem).
2. 6/AC = 4/3, so AC = (6 × 3)/4 = 4.5 cm.

Teacher's Note:
1. The angle bisector divides the opposite side in the ratio of the other two sides.
2. Place the ratio so that AB pairs with BD.

Answer: (c) 1 : 2

Explanation:
1. Let AB = BC = x. Then AC² = x² + x² = 2x², so AC = x√2.
2. For similar triangles, the ratio of areas is the ratio of the squares of corresponding sides: area ABE/area ACD = AB²/AC² = x²/2x² = 1/2.
3. The ratio is 1 : 2.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides.
2. AB corresponds to AC in the two similar triangles.

Answer: (c) 5 : 6

Explanation:
1. Equal corresponding angles make the triangles similar.
2. The ratio of areas equals the ratio of the squares of corresponding heights: h₁²/h₂² = 25/36.
3. So h₁/h₂ = 5/6.

Teacher's Note:
1. For similar triangles the ratio of areas is the square of the ratio of corresponding altitudes.
2. Take the square root of the area ratio.

Answer: (a) 30°

Explanation:
1. ∠A = 180° − (70° + 50°) = 60°.
2. AB/AC = BD/DC means AD is the bisector of ∠A.
3. ∠BAD = 60°/2 = 30°.

Teacher's Note:
1. The relation AB/AC = BD/DC is the angle bisector theorem and its converse.
2. Find ∠A first with the angle sum property.

Answer: (a) 9

Explanation:
1. Area ABC / area DEF = (BC/EF)² = (2.1/2.8)² = (3/4)² = 9/16.
2. Area ABC = 9/16 × 16 = 9 cm².

Teacher's Note:
1. Match corresponding sides BC and EF.
2. Square the ratio of sides to get the ratio of areas.

Answer: (a) 25 cm and 7 cm

Explanation:
1. In right triangle ACB: AB² = AC² + CB² = 20² + 15² = 625, so AB = 25 cm.
2. In right triangle ABD: BD² = AB² − AD² = 625 − 576 = 49, so BD = 7 cm.

Teacher's Note:
1. Use the Pythagoras theorem twice, first in ΔACB and then in ΔABD.
2. Identify the right angles from the figure.

Answer: (b) DF × EF = FB × FA

Explanation:
1. Since AD ∥ BE, ∠1 = ∠2 (vertically opposite angles) and ∠3 = ∠4 (alternate angles).
2. So ΔFBE ~ ΔFDA by AA similarity.
3. Hence EF/FA = FB/DF, which gives DF × EF = FB × FA. Option (a) has AB in place of DF, so it is not correct.

Teacher's Note:
1. Name the similar triangles by matching equal angles in order.
2. Cross-multiply the proportion to match the option form.

Answer: (a) 6 − \( \frac{3y}{4} \)

Explanation:
1. Draw AL ⊥ BC. L is the midpoint of BC, so BL = 6 and AL = √(100 − 36) = 8 cm.
2. The rectangle is symmetric about AL, so BS = 6 − x. Also PS ∥ AL, so ΔBSP ~ ΔBLA.
3. PS/AL = BS/BL gives y/8 = (6 − x)/6, so x = 6 − 3y/4.

Teacher's Note:
1. Use the symmetry of the isosceles triangle about the altitude AL.
2. PS is parallel to AL, which gives similar triangles.

Answer: (d) 4 : 1

Explanation:
1. All equilateral triangles are similar, so ΔABC ~ ΔEBD.
2. BC = 2BD, so area ABC / area BDE = BC²/BD² = (2BD)²/BD² = 4/1.
3. The ratio is 4 : 1.

Teacher's Note:
1. Equilateral triangles are always similar.
2. Doubling a side makes the area four times.

Answer: (a) 14 sq.cm

Explanation:
1. Area ABC / area DEF = AB²/DE² = (2DE)²/DE² = 4.
2. Area DEF = 56/4 = 14 cm².

Teacher's Note:
1. Side ratio 2 means area ratio 4.
2. Divide the larger area by 4.

Answer: (b) 18 cm

Explanation:
1. Let the legs be a and b and the hypotenuse c. Then ab = 80, a + b + c = 40 and a² + b² = c².
2. (a + b)² = (40 − c)² = a² + b² + 2ab = c² + 160.
3. 1600 − 80c + c² = c² + 160, so 80c = 1440 and c = 18 cm.

Teacher's Note:
1. Use (a + b)² = a² + b² + 2ab to bring in the area.
2. The c² terms cancel, leaving a simple equation.

Answer: (c) \( 1 : \sqrt{2} \)

Explanation:
1. PQ ∥ BC, so ΔAPQ ~ ΔABC.
2. Area ABC = 2 × area APQ, so area APQ / area ABC = 1/2.
3. (AP/AB)² = 1/2, so AP : AB = 1 : √2.

Teacher's Note:
1. Equal areas of the two parts means the small triangle is half of the whole triangle.
2. Take the square root of the area ratio to get the side ratio.

Answer: (a) 9

Explanation:
1. ar(PQR)/ar(ABC) = (QR/BC)² = 3² = 9.
2. Here QR/BC is the reciprocal of BC/QR = 1/3, which is 3.

Teacher's Note:
1. Look at which triangle is in the numerator.
2. Reverse the given ratio before squaring.

Answer: (b) 18225 m²

Explanation:
1. Let each equal side be s. Then s² + s² = 270², so s² = 270²/2 = 36450.
2. Area = ½ × s × s = ½ × 36450 = 18225 m².

Teacher's Note:
1. Area of an isosceles right triangle = ½ s², which needs only s².
2. No square root is required.

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 = 3/2 × 10 = 15 cm.

Teacher's Note:
1. Perimeters of similar triangles are in the same ratio as their sides.
2. Corresponding sides are AB and PQ.

Answer: (c) 2 cm

Explanation:
1. DE ∥ BC, so AD/DB = AE/EC (basic proportionality theorem).
2. 1.5/3 = 1/EC, so EC = 2 cm.

Teacher's Note:
1. Use the basic proportionality theorem with the parts on the two sides.
2. Keep the order: upper part over lower part on both sides.

Answer: (b) \( \frac{ac}{b + c} \)

Explanation:
1. ∠LMK = ∠PNK = 46° (corresponding angles), and ∠K is common.
2. So ΔKNP ~ ΔKML by AA similarity.
3. KN/KM = NP/ML, that is c/(b + c) = x/a, so x = ac/(b + c).

Teacher's Note:
1. Name the triangles with the common angle K first.
2. KM = MN + NK = b + c.

Answer: (a) All isosceles triangles are similar.

Explanation:
1. Isosceles triangles can have different angles, so they need not be similar. Statement (a) is false.
2. Equilateral triangles have all angles 60°, so they are similar. Any two circles have the same shape, so they are similar.

Teacher's Note:
1. Similar figures have the same shape; equal angles and proportional sides are needed.
2. Look for a counter-example for the false statement.

Answer: (b) 1 : 2

Explanation:
1. ar(ABC)/ar(APQ) = BC²/PQ² = 1/4.
2. So BC/PQ = 1/2, that is 1 : 2.

Teacher's Note:
1. Write the ratio with ABC in the numerator.
2. Take the square root to get the sides.

Answer: (d) 6 cm

Explanation:
1. 9/16 = BC²/QR² = (4.5)²/QR².
2. QR² = 16 × 20.25/9 = 36, so QR = 6 cm.

Teacher's Note:
1. Area ratio 9 : 16 means side ratio 3 : 4.
2. BC = 4.5 corresponds to 3 parts, so each part is 1.5 cm.

Answer: (a) \( 8\sqrt{2} \) cm

Explanation:
1. Let the side be x. By the Pythagoras theorem, x² + x² = 16² = 256.
2. x² = 128, so x = 8√2 cm.

Teacher's Note:
1. The diagonal of a square is the hypotenuse of an isosceles right triangle.
2. Side = diagonal/√2 = 16/√2 = 8√2.

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

Explanation:
1. AD ⊥ BC in an equilateral triangle, so D is the midpoint of BC and BD = p.
2. In ΔADB: AD² = AB² − BD² = 4p² − p² = 3p², so AD = √3 p.

Teacher's Note:
1. The altitude of an equilateral triangle bisects the base.
2. Altitude = (√3/2) × side.

Answer: (c) 9 : 7

Explanation:
1. The ratio of areas equals the square of the ratio of corresponding medians.
2. (m₁/m₂)² = 81/49, so m₁/m₂ = 9/7.

Teacher's Note:
1. Medians, altitudes and sides of similar triangles are all in the same ratio.
2. Square root of the area ratio gives the length ratio.

Answer: (c) \( \frac{\sqrt{5} - 1}{2} \)

Explanation:
1. AC = BC, so ∠A = ∠B = 72° and ∠C = 36°. AD bisects ∠A, so ∠DAB = 36°.
2. In ΔADB, ∠ADB = 180° − 36° − 72° = 72° = ∠B, so AB = AD. Also ∠CAD = 36° = ∠C, so AD = CD = 1, hence AB = 1 and AC = BC = 1 + x where BD = x.
3. Angle bisector theorem: AC/AB = CD/BD gives (1 + x)/1 = 1/x, so x² + x − 1 = 0.
4. x = (−1 + √5)/2 (the positive root).

Teacher's Note:
1. Find all angles first; two isosceles triangles appear.
2. The angle bisector theorem gives a quadratic in BD.

Answer: (c) \( 14 \times 3^{3/4} \) cm

Explanation:
1. In ΔABC and ΔDAC: ∠BAC = ∠ADC and ∠C is common, so ΔABC ~ ΔDAC.
2. BC/AC = AC/DC, so BC × DC = AC² = 441 = area of the rectangle.
3. Equilateral triangle of side s: (√3/4)s² = 441, so s² = 1764/√3 and s = 42/3^{1/4} = 14 × 3^{3/4}.

Teacher's Note:
1. First show that BC × DC = AC² using similar triangles.
2. Set the equilateral triangle's area (√3/4)s² equal to 441.

Answer: (a) 12

Explanation:
1. AC = √(25² − 15²) = 20.
2. Area of ΔABC: ½ × BC × AD = ½ × AB × AC, so 25 × AD = 15 × 20 and AD = 12.
3. AEDF is a rectangle (three right angles at A, E, F), so its diagonals are equal: EF = AD = 12.

Teacher's Note:
1. A quadrilateral with three right angles is a rectangle.
2. The diagonals of a rectangle are equal.

Answer: (a) A and C

Explanation:
1. D is the midpoint of BC, so BD = a/2.
2. AD² = AB² − BD² = a² − a²/4 = 3a²/4. This is A.
3. Since BC = a, 3a²/4 = (3/4)BC², which is C. D is AD itself, not AD², and B is wrong.

Teacher's Note:
1. The altitude is (√3/2)a, so its square is 3a²/4.
2. BC and a are the same length here.

Answer: (d) If the areas of two similar triangles are equal, then the triangles are congruent.

Explanation:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides, altitudes or medians, so (a), (b) and (c) are wrong.
2. If the areas are equal the ratio of squares is 1, so the corresponding sides are equal and the similar triangles are congruent. (d) is correct.

Teacher's Note:
1. Similarity gives a ratio of areas equal to the square of the ratio of sides.
2. Equal areas force a ratio of 1.

Answer: (a) I

Explanation:
1. I: ratio of areas = (2/1)² = 4 : 1, which is true.
2. II: AP : AB = 1 : 3, so the area ratio is 1/9, not 1/4. False.
3. III: the ratio of sides = √(9/16) = 3 : 4, not 3 : 16. False.
4. Only I is correct.

Teacher's Note:
1. Square the ratio to go from sides to areas, and take the root to go back.
2. In II use AP : AB (1 : 3), not AP : PB.

Answer: (a) Length of greatest side is 10 cm

Explanation:
1. The greatest side is the hypotenuse AC = √(8² + 6²) = √100 = 10 cm. So (a) is correct.
2. The legs are unequal, so neither acute angle is 45°, and the Pythagoras theorem applies to every right triangle.

Teacher's Note:
1. The longest side of a right triangle is the hypotenuse.
2. Equal acute angles of 45° need equal legs.

Answer: (a) their corresponding angles are equal.

Explanation:
1. Two triangles are similar when their corresponding angles are equal (and then the corresponding sides are proportional).
2. Equal sides would make them congruent, and two right triangles can have different acute angles.

Teacher's Note:
1. Similar means the same shape, not necessarily the same size.
2. AA is enough to prove similarity.

Answer: (d) (d)

Explanation:
1. (a): proportional division of the diagonals gives a pair of parallel sides, so the quadrilateral is a trapezium. True.
2. (b): by the mid-point theorem both pairs of opposite sides of the figure are parallel to a diagonal, so it is a parallelogram. True.
3. (c): medians of similar triangles are in the same ratio as the sides. True.
4. Since (a), (b) and (c) are all correct, none of them is incorrect: option (d).

Teacher's Note:
1. Check each statement before choosing.
2. When all listed statements are true, 'None of the above' is the answer.

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

Explanation:
1. By the basic proportionality theorem (Thales), AD/DB = AE/EC.
2. Taking reciprocals and adding 1: DB/AD + 1 = EC/AE + 1, that is (AD + DB)/AD = (AE + EC)/AE, so AB/AD = AC/AE. The Assertion is true.
3. The Reason is the basic proportionality theorem used to prove it, so it explains the Assertion: option (a).

Teacher's Note:
1. The Reason is Thales' theorem, the basis of the Assertion.
2. Add 1 to both sides to move from parts to the whole side.

Answer: (d) A is false, but R is true.

Explanation:
1. AC = BC and ∠C = 90°, so AB² = AC² + BC² = 2AC², not 3AC². The Assertion is false.
2. The Reason is the converse of the Pythagoras theorem: AB² = AC² + BC², so ∠C = 90°. The Reason is true.
3. Assertion false, Reason true: option (d).

Teacher's Note:
1. Apply the Pythagoras theorem in the right triangle.
2. The Reason is the converse of the Pythagoras theorem.

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

Explanation:
1. ar(ABC)/ar(DEF) = BC²/EF² = 16/25, so ar(DEF) = 64 × 25/16 = 100 cm². The Assertion is true.
2. The Reason is also a true standard result, but the Assertion was found using the corresponding sides, not altitudes, so the Reason is not the explanation: option (b).

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of any pair of corresponding linear measures.
2. The Assertion uses sides, while the Reason talks about altitudes.