Heron's Formula — Chapter Test | Class 9 IMO

Q1

The sides of a triangle are 5 cm, 12 cm, and 13 cm. The altitude drawn on the longest side is:

Area = ½ × 5 × 12 = 30 cm². Also, ½ × 13 × altitude = 30 → altitude = 60/13 cm.

Q2

If each side of a triangle is increased by 100%, the percentage increase in its area is:

Increasing by 100% means the sides are doubled. New area = 4 × Original Area. Increase = 4A − A = 3A, which is 300%.

Q3

Find the area of a triangle whose sides are 50 cm, 65 cm, and 65 cm.

s = (50+65+65)/2 = 90. Area = √(90 × 40 × 25 × 25) = √2250000 = 1500 cm².

Q4

Find the length of the altitude corresponding to the smallest side of a triangle whose sides are 17 cm, 25 cm, and 26 cm.

s = 34. Area = √(34 × 17 × 9 × 8) = 204 cm². The smallest side is 17 cm. ½ × 17 × h = 204 → h = 408 / 17 = 24 cm.

Q5

A quadrilateral field has a diagonal of 24 m. The lengths of perpendiculars dropped on it from the opposite vertices are 8 m and 13 m. The area of the field is:

Area = ½ × d × (h₁ + h₂) = ½ × 24 × (8 + 13) = 12 × 21 = 252 m².

Q6

Find the area of a triangle with sides 11 cm, 60 cm, and 61 cm.

These sides form a right triangle (11² + 60² = 121 + 3600 = 3721 = 61²). Area = ½ × 11 × 60 = 330 cm². (Also verifiable via Heron's: s=66, Area = √(66×55×6×5) = 330).

Q7

A floral design is made from 16 identical triangular tiles and has a total area of 576√6 cm². If polishing costs 50 paise per cm², what is the total polishing cost?

Total area = 576√6 cm². Cost in Rupees = 576√6 × 0.50 = ₹288√6.

Q8

A quadrilateral park ABCD has AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m and angle C = 90°. Its total area is:

Join BD. In right ΔBCD, BD = √(12² + 5²) = 13 m. Area BCD = ½ × 12 × 5 = 30 m². For ΔABD (sides 9, 8, 13), s = 15. Area = √(15 × 6 × 7 × 2) = 6√35 m². Total = 30 + 6√35 m².

Q9

The sides of a triangle are 15 cm, 14 cm, and 13 cm. Calculate the radius of its inscribed circle (inradius).

s = 21 cm. Area A = √(21 × 6 × 7 × 8) = 84 cm². The inradius r = A / s = 84 / 21 = 4 cm.

Q10

Find the area of a convex quadrilateral with sides 7 cm, 24 cm, 20 cm, and 15 cm, given that the angle between the first two sides is 90°.

The right triangle (7, 24) has diagonal = √(7²+24²) = 25 cm and area = 84 cm². The second triangle has sides 15, 20, 25 (which is also a right triangle), area = 150 cm². Total = 234 cm².

Q11

Find the area of a triangle having sides 13 cm, 14 cm, and 15 cm.

s = (13+14+15)/2 = 21 cm. Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm².

Q12

A triangular plot with sides 18 m, 24 m, and 30 m needs clearing. If the contractor charges ₹10 per m², what is the total bill?

The sides form a right triangle (18, 24, 30 are a multiple of 3, 4, 5). Area = ½ × 18 × 24 = 216 m². Bill = 216 × 10 = ₹2160.

Q13

If every side of a triangle is doubled, the ratio of the area of the original triangle to the new triangle is:

When sides are doubled, the semi-perimeter doubles, making each (s-a) term double. √(2s × 2(s-a) × 2(s-b) × 2(s-c)) = √16 × Area = 4 × Area. Ratio is 1:4.

Q14

The area of a triangle is calculated using base = 12 cm and height = 8 cm. If calculated correctly using Heron's formula with its three side lengths, the result will be:

Both methods yield the exact same area. Area = ½ × base × height = ½ × 12 × 8 = 48 cm².

Q15

Calculate the area of quadrilateral ABCD where AB = 3 cm, BC = 4 cm, CD = 5 cm, DA = 5 cm, and AC = 5 cm.

ΔABC is a right triangle (3,4,5), area = ½ × 3 × 4 = 6. ΔACD has sides 5, 5, 5 (equilateral), area = (√3/4) × 5² = 25√3/4. Total = 6 + 25√3/4 cm².