CBSE Class 9 Mathematics Areas of Parallelograms and Triangles MCQs Set 13

Answer: (a) 6.4 cm

Explanation:
1. Area = base × corresponding altitude = 8 × 4 = 32 cm².
2. The same area equals AD × 5, so AD = 32 ÷ 5 = 6.4 cm.

Teacher's Note:
1. Area is the same whichever base is used.
2. Longer base goes with the shorter altitude.

Answer: (b) 12.8 cm

Explanation:
1. Taking DC = AB = 16 cm as base, area = 16 × 8 = 128 cm².
2. Taking AD as base with altitude CF = 10 cm, 128 = AD × 10, so AD = 12.8 cm.

Teacher's Note:
1. Opposite sides of a parallelogram are equal, so DC = AB.

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

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

Teacher's Note:
1. Equate the two expressions for the same area.

Answer: (d) \( \dfrac16 \) × ar(△ABC)

Explanation:
1. A median divides the triangle into two equal parts: ar(△ABD) = ½ ar(△ABC).
2. AP = ⅓AD, and triangles with the same vertex B and bases on AD are in the ratio of the bases, so ar(△ABP) = ⅓ ar(△ABD) = ⅙ ar(△ABC).

Teacher's Note:
1. Triangles with a common vertex and bases on the same line have areas in the ratio of the bases.

Answer: (a) 5 : 8

Explanation:
1. △ADC and △ABC have the same altitude from A, so their areas are in the ratio of the bases DC : BC.
2. DC : BC = 5 : 8.

Teacher's Note:
1. BC = BD + DC = 3 + 5 = 8 parts.

Answer: (b) 192 cm²

Explanation:
1. Area = ½ × d₁ × d₂ = ½ × 16 × 24 = 192 cm².

Teacher's Note:
1. The diagonals of a rhombus divide it into four congruent right triangles.

Answer: (c) 114 cm²

Explanation:
1. BD = √(17² − 8²) = 15 cm and AB = √(15² − 12²) = 9 cm.
2. Area = ½ × 12 × 9 + ½ × 15 × 8 = 54 + 60 = 114 cm².

Teacher's Note:
1. Split the quadrilateral along the diagonal BD into two right triangles.

Answer: (d) 13 cm, 40 cm²

Explanation:
1. Each slant side makes a right triangle with height 4 and hypotenuse 5, so its base is 3 cm.
2. DC = 7 + 3 + 3 = 13 cm. Area = ½ × (7 + 13) × 4 = 40 cm².

Teacher's Note:
1. Drop perpendiculars from A and B to DC.

Answer: (a) 150 cm²

Explanation:
1. ar(ABCD) = ar(△ABD) + ar(△CBD) = ½ × 20 × 10 + ½ × 20 × 5.
2. = 100 + 50 = 150 cm².

Teacher's Note:
1. Take BD as the common base of both triangles.

Answer: (b) 40 cm²

Explanation:
1. △BOX ≅ △DOY (ASA), because BO = DO and AB ∥ DC. Hence the part cut off on one side of XY can be exchanged with the other, and XY divides the parallelogram into two parts of equal area.
2. So ar(AXYD) = ½ × 80 = 40 cm².

Teacher's Note:
1. Any line through the centre of a parallelogram halves its area.

Answer: (c) 1 : 1

Explanation:
1. ar(△ABD) = ar(△ACD) and ar(△EBD) = ar(△ECD), both because the base is halved by D.
2. Subtract: ar(△ABE) = ar(△ACE).

Teacher's Note:
1. Medians split triangles into equal areas, also with any point on the median.

Answer: (d) bisects AD

Explanation:
1. Let AD meet BC at O. Equal areas on the same base give equal altitudes from A and D to BC.
2. Then △AOL ≅ △DOM (AAS), so AO = DO.

Teacher's Note:
1. Equal altitudes on opposite sides make O the midpoint of AD.

Answer: (a) 12 cm²

Explanation:
1. ar(△ABD) = ½ ar(△ABC) = 24 cm².
2. E is the midpoint of AD, so ar(△BED) = ½ ar(△ABD) = 12 cm².

Teacher's Note:
1. A median halves the area each time.

Answer: (b) 25 cm²

Explanation:
1. The midpoints divide △ABC into four congruent triangles of 12.5 cm² each.
2. Parallelogram BDEF consists of two of them: 25 cm².

Teacher's Note:
1. Midpoint theorem makes BDEF a parallelogram.

Answer: (c) \( 16\sqrt3 \) cm²

Explanation:
1. The altitude splits the base into 4 cm and 4 cm, so the height is √(64 − 16) = 4√3 cm.
2. Area = ½ × 8 × 4√3 = 16√3 cm², i.e. (√3/4)a² with a = 8.

Teacher's Note:
1. Formula: area = (√3/4) × side².

Answer: (d) 48 cm²

Explanation:
1. The heights of △APB and △PCD from P add up to AD = 8 cm.
2. Sum = ½ × 12 × 8 = 48 cm², half of the rectangle.

Teacher's Note:
1. It does not depend on where P is.

Answer: (a) 6 cm²

Explanation:
1. Using ar(△APB) × ar(△CPD) = ar(△APD) × ar(△BPC): 4 × 9 = 6 × ar(△BPC).
2. ar(△BPC) = 36 ÷ 6 = 6 cm².

Teacher's Note:
1. The product of areas of opposite triangles is equal.

Answer: (b) 60 cm²

Explanation:
1. Parallelograms on the same base and between the same parallels have equal areas.

Teacher's Note:
1. Same base and same height imply the same area.

Answer: (c) 20 cm²

Explanation:
1. Triangles ABD, ADE, AEC have equal bases and the same altitude from A, so they have equal areas, each one third of 60 cm².

Teacher's Note:
1. Equal bases and common height give equal areas.

Answer: (d) 2 : 5

Explanation:
1. Both triangles have the same height from A, so areas are in the ratio of the bases BD : DC = 2 : 5.

Teacher's Note:
1. Common vertex gives area ratio equal to base ratio.

Answer: (a) 8

Explanation:
1. Sum of exterior angles = 360°, so interior sum = 1080° = (n − 2) × 180°.
2. n − 2 = 6, n = 8.

Teacher's Note:
1. Exterior angles of any convex polygon add to 360°.