CBSE Class 10 Mathematics Triangles MCQs Set 21

Answer: (b) 4

Explanation:
1. By the basic proportionality theorem, AD/DB = AE/EC.
2. x/(x - 2) = (x + 2)/(x - 1), so x(x - 1) = (x + 2)(x - 2).
3. x² - x = x² - 4, so x = 4.

Teacher's Note:
1. Apply the basic proportionality theorem.
2. Check: 4/2 = 6/3 = 2.

Answer: (a) 2.8 cm

Explanation:
1. BD = BC - DC = 4 cm.
2. By the angle bisector theorem AB/AC = BD/DC = 4/6.
3. AB = 4.2 × 4/6 = 2.8 cm.

Teacher's Note:
1. Find BD first.
2. The bisector divides the opposite side in the ratio of the other two sides.

Answer: (c) 130°

Explanation:
1. Angle C = 180° - 80° - 60° = 40°.
2. In triangle BDC, angle DBC = 30° and angle DCB = 20°.
3. Angle BDC = 180° - 50° = 130° (this is 90° + A/2).

Teacher's Note:
1. Find angle C first.
2. The angle between two internal bisectors is 90° + half of the third angle.

Answer: (b) \( a^2 + b^2 + c^2 - s^2 \)

Explanation:
1. Expand: 3s² - 2s(a + b + c) + a² + b² + c².
2. Since a + b + c = 2s: 3s² - 4s² + a² + b² + c² = a² + b² + c² - s².

Teacher's Note:
1. Expand each square.
2. Substitute a + b + c = 2s.

Answer: (b) 6 cm

Explanation:
1. CD = 12 - 9 = 3 cm.
2. Triangles CAD and CBA have a common angle C and angle ADC = angle BAC, so they are similar.
3. AC/BC = CD/CA, so AC² = CD × BC = 3 × 12 = 36 and AC = 6 cm.

Teacher's Note:
1. Prove the triangles similar first.
2. Match the corresponding sides carefully.

Answer: (d) 4.5 cm

Explanation:
1. The centroid divides each median in the ratio 2 : 1, so AG = 2GD = 3 cm.
2. AD = AG + GD = 3 + 1.5 = 4.5 cm.

Teacher's Note:
1. The centroid is two-thirds of the way along each median from the vertex.
2. AD = 3 × GD.

Answer: (a) 20 cm

Explanation:
1. For an equilateral triangle of side s, the circumradius R = s/√3.
2. R = 20√3/√3 = 20 cm.

Teacher's Note:
1. The circumcentre is also the centroid.
2. R = (2/3) × height = (2/3)(s√3/2) = s/√3.

Answer: (c) 1 : 4

Explanation:
1. Triangles ANC and BAC are similar (angle C common, right angles at N and A).
2. Ratio of areas = (AC/BC)² = (6/12)² = 1/4.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides.
2. AC in triangle ANC corresponds to BC in triangle BAC.

Answer: (b) \( R^2 \)

Explanation:
1. The base is the diameter 2R and the height is at most R (at the top of the semicircle).
2. Largest area = ½ × 2R × R = R².

Teacher's Note:
1. The height cannot exceed the radius.
2. The largest triangle is isosceles on the diameter.

Answer: (b) 180° + angle BAC

Explanation:
1. Exterior angle at B = 180° - B and at C = 180° - C.
2. Sum = 360° - (B + C) = 360° - (180° - A) = 180° + A.

Teacher's Note:
1. The exterior angle is supplementary to the interior angle.
2. B + C = 180° - A.

Answer: (c) right angled

Explanation:
1. x + 3x + y = 180°, so y = 180° - 4x.
2. 3(180° - 4x) - 5x = 30 gives 540° - 17x = 30, so x = 30°.
3. A = 30°, B = 90°, C = 60°, so the triangle is right angled.

Teacher's Note:
1. Use the angle sum property.
2. Solve the two equations in x and y.

Answer: (c) 130°

Explanation:
1. Angle BOC = 180° - (B/2 + C/2) = 180° - (180° - 80°)/2.
2. = 180° - 50° = 130°.

Teacher's Note:
1. B + C = 100°.
2. Half of that is 50°.

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

Explanation:
1. Triangle XBY ~ triangle ABC and ar(XBY) = ½ ar(ABC).
2. (XB/AB)² = 1/2, so XB/AB = 1/√2.
3. AB : XB = √2 : 1.

Teacher's Note:
1. Equal parts means the smaller triangle has half the area.
2. Take square roots of the area ratio.

Answer: (a) 2 : 1

Explanation:
1. ar(XBY) is half of ar(ABC).
2. So ar(ABC) : ar(XBY) = 2 : 1.

Teacher's Note:
1. The line splits the area into two equal parts.
2. The smaller triangle is half of the whole.

Answer: (c) 4.2 cm

Explanation:
1. Ratio of areas = 12 : 48 = 1 : 4, so the ratio of corresponding heights is 1 : 2.
2. Height of the bigger triangle = 2 × 2.1 = 4.2 cm.

Teacher's Note:
1. The ratio of areas is the square of the ratio of corresponding altitudes.
2. Take the square root.

Answer: (b) 60°

Explanation:
1. Angle C = 60°, so angle ACE = 30° after the bisector.
2. In right triangle CED (right angle at D), angle ECD = 30°.
3. Angle CED = 90° - 30° = 60°.

Teacher's Note:
1. Find angle C first.
2. Use the angle sum in the right triangle CED.

Answer: (c) \( \frac{x}{\sqrt{2}} \)

Explanation:
1. Angle B = 45° and angle A = 90°, so triangle ABD is a right triangle with angle ABD = 45°.
2. AD = AB sin 45° = x/√2.

Teacher's Note:
1. Triangle ABD is a 45°-45°-90° triangle.
2. The hypotenuse AB is √2 times AD.

Answer: (b) 90°

Explanation:
1. AD/BD = DC/AD with the right angles at D, so triangle ADB ~ triangle CDA (SAS).
2. Angle BAD = angle ACD, so angle BAD + angle DAC = angle ACD + angle DAC = 90°.
3. Angle BAC = 90°.

Teacher's Note:
1. Show the two small triangles are similar.
2. Use the angle sum in the right triangle ADC.

Answer: (a) 294 cm²

Explanation:
1. Triangles POB and QOA are similar (right angles and vertically opposite angles at O).
2. Ratio of sides PO : QO = 5 : 7, so the area ratio is 25 : 49.
3. Area QOA = 150 × 49/25 = 294 cm².

Teacher's Note:
1. Prove similarity by AA.
2. The area ratio is the square of the side ratio.

Answer: (c) 4 : 9

Explanation:
1. The ratio of altitudes is 6 : 9 = 2 : 3.
2. The ratio of areas is (2/3)² = 4 : 9.

Teacher's Note:
1. Areas are in the ratio of the squares of corresponding altitudes.
2. Simplify 6 : 9 first.

Answer: (a) \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \)

Explanation:
1. Area = ½ab = ½·c·p, so p = ab/c with c² = a² + b².
2. p² = a²b²/(a² + b²).
3. Hence 1/p² = (a² + b²)/(a²b²) = 1/b² + 1/a².

Teacher's Note:
1. Find the area in two ways.
2. Take the reciprocal of p².

Answer: (a) 5.6 cm

Explanation:
1. The scale factor is BC/PQ = 10/5 = 2.
2. CA/AP = 2, so CA = 2 × 2.8 = 5.6 cm.

Teacher's Note:
1. Match the corresponding sides: CB with PQ and CA with PA.
2. Find the scale factor first.

Answer: (a) 16.96 cm²

Explanation:
1. s = (5.6 + 6.5 + 10)/2 = 11.05.
2. Area = √(11.05 × 5.45 × 4.55 × 1.05) = √287.7 = 16.96 cm² (to two decimals).

Teacher's Note:
1. Use Heron's formula.
2. Compute s - a, s - b and s - c carefully.

Answer: (b) 4 : 1

Explanation:
1. The sides are in the ratio 2 : 1.
2. The areas are in the ratio 2² : 1² = 4 : 1.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of the sides.
2. Use the scale factor 2.