CBSE Class 9 Mathematics Areas of Parallelograms and Triangles MCQs Set 14

Answer: (c) \(\dfrac{108}{11}\) cm

Explanation:
1. Area = AB × 9 = 12 × 9 = 108 cm².
2. Also area = AD × 11, so AD = 108 ÷ 11 cm.

Teacher's Note:
1. Area of a parallelogram is the same with either base.
2. Equate base × height for both bases.

Answer: (d) \(\dfrac16\) × area of triangle ABC

Explanation:
1. A median divides a triangle into two triangles of equal area, so ar(ABD) = (1/2) ar(ABC).
2. Triangles ABP and ABD have the same height from B, with bases in the ratio AP : AD = 1 : 3, so ar(ABP) = (1/3) ar(ABD).
3. Hence ar(ABP) = (1/3) × (1/2) ar(ABC) = (1/6) ar(ABC).

Teacher's Note:
1. Use the median first, then the ratio of bases.
2. Triangles with the same height have areas in the ratio of their bases.

Answer: (c) 5 : 8

Explanation:
1. Triangles ADC and ABC have the same height from A.
2. Their areas are in the ratio of the bases DC : BC = 5 : 8.

Teacher's Note:
1. Same height means area ratio equals base ratio.
2. DC is 5 parts out of 8 parts.

Answer: (d) 192 cm²

Explanation:
1. Area = (1/2) × d₁ × d₂ = (1/2) × 16 × 24 = 192 cm².

Teacher's Note:
1. Area of a rhombus is half the product of the diagonals.
2. Do not forget the half.

Answer: (a) 56 cm²

Explanation:
1. Area = (1/2) × (8 + 6) × 8 = 56 cm².

Teacher's Note:
1. Area of trapezium = half the sum of parallel sides × distance between them.
2. Here 14 × 8 ÷ 2 = 56.

Answer: (d) 40 cm²

Explanation:
1. Drop perpendiculars from A and B to DC. Each foot is \(\sqrt{5^2 - 4^2} = 3\) cm from D and C.
2. DC = 7 + 3 + 3 = 13 cm.
3. Area = (1/2) × (7 + 13) × 4 = 40 cm².

Teacher's Note:
1. Isosceles trapezium: the two overhangs are equal.
2. Pythagoras gives the overhang as 3 cm.

Answer: (a) 150 cm²

Explanation:
1. Area = ar(ABD) + ar(CBD) = (1/2) × BD × AL + (1/2) × BD × CM.
2. = (1/2) × 20 × (10 + 5) = 150 cm².

Teacher's Note:
1. Split the quadrilateral along the diagonal BD.
2. Add the two triangle areas.