Areas of Parallelograms & Triangles — Chapter Test | Class 9 IMO

Q1

Assertion (A): Median of a triangle divides it into two triangles of equal areas. Reason (R): The centroid of a triangle divides the median in the ratio 2:1.

The median divides the base into two equal parts, so the two triangles have equal bases and the same height, giving equal areas. The centroid ratio (2:1) is true but doesn't explain the area division by the median.

Q2

In a trapezium-shaped garden, the parallel sides are 20 m and 30 m long, and the perpendicular distance between them is 15 m. Find the area of the garden.

Area = ½ × (20 + 30) × 15 = ½ × 50 × 15 = 25 × 15 = 375 m².

Q3

A triangular hoarding is put up by a shopkeeper in Delhi. It has sides 13 m, 14 m, and 15 m. To paint it at ₹10 per m², what will be the total cost?

s = (13+14+15)/2 = 21. Area = √(21×8×7×6) = √7056 = 84 m². Cost = 84 × 10 = ₹840.

Q4

P is any point on the median AD of triangle ABC. The ratio of Area(ABP) to Area(ACP) is:

AD is the median, so triangles ABD and ACD have equal areas. PD is the median of PBC, so PBD and PCD have equal areas. Subtracting gives Area(ABP) = Area(ACP). Ratio is 1:1.

Q5

A quadrilateral ABCD has diagonal AC = 15 cm. The perpendiculars drawn from B and D to AC are 4 cm and 6 cm respectively. The area of the quadrilateral is:

Area = Area(ABC) + Area(ADC) = ½ × 15 × 4 + ½ × 15 × 6 = ½ × 15 × (4 + 6) = ½ × 150 = 75 cm².

Q6

ABCD is a parallelogram and P is any point in its interior. Then Area(APB) + Area(PCD) is equal to:

Drawing a line through P parallel to the base shows that Area(APB) + Area(PCD) = ½ Area(ABCD).

Q7

Two adjacent sides of a parallelogram are 5 cm and 4 cm. If the diagonal connecting the ends of these sides is 7 cm, the area of the parallelogram is:

Sides of triangle: 5, 4, 7. s = (5+4+7)/2 = 8. Area of triangle = √(8×3×4×1) = √96 = 4√6 cm². Area of parallelogram = 2 × 4√6 = 8√6 cm².

Q8

In a parallelogram ABCD, AB = 10 cm. The altitudes corresponding to the sides AB and AD are respectively 6 cm and 8 cm. Find AD.

Area = AB × h₁ = AD × h₂. 10 × 6 = AD × 8. 60 = 8 × AD. AD = 60/8 = 7.5 cm.

Q9

In a triangle ABC, AB = 6 cm, BC = 8 cm and AC = 10 cm. A perpendicular dropped from B meets AC at D. The length of BD is:

It's a right triangle at B since 6² + 8² = 10². Area = ½ × 6 × 8 = 24 cm². Area also = ½ × AC × BD = ½ × 10 × BD. 5 × BD = 24. BD = 4.8 cm.

Q10

To find the area of a quadrilateral field ABCD, it is divided by diagonal AC which is 24 m. Perpendicular offsets from B and D are 8 m and 12 m. The area of the field is:

Area = ½ × d × (h₁ + h₂) = ½ × 24 × (8 + 12) = 12 × 20 = 240 m².

Q11

Which of the following transformations changes the area of a triangle?

Translation, rotation, and reflection are rigid transformations that preserve shape and size (isometries). Scaling changes the dimensions and therefore the area.

Q12

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Which of the following is true?

Triangles ABD and ABC are on same base AB and between same parallels. Area(ABD) = Area(ABC). Subtracting Area(AOB) gives Area(AOD) = Area(BOC).

Q13

In a rhombus ABCD, diagonals AC and BD intersect at O. If Area(AOB) = 15 cm², what is the area of the rhombus?

Diagonals of a rhombus divide it into four congruent right-angled triangles. Area = 4 × 15 = 60 cm².

Q14

The area of a trapezium whose parallel sides are 9 cm and 16 cm and the distance between them is 8 cm is:

Area = ½ × (Sum of parallel sides) × height = ½ × (9 + 16) × 8 = ½ × 25 × 8 = 100 cm².

Q15

G is the centroid of triangle ABC. The area of triangle GAB is 15 cm². The area of triangle ABC is:

The centroid connects to the vertices to form 3 smaller triangles of equal area: GAB, GBC, GCA. Total Area = 3 × 15 = 45 cm².