#### Class Nine Chap 12 Heron Formula Math Extra Questions

## Chapter 12: Heron’s Formula

**Short Answer Questions Type [2 Marks] **

**1. The sides of a triangle are in the ratio 3: 4: 5. If the perimeter of the triangle is 84 cm, then find the area of the triangle. **

**Sol. **We have, 3x + 4x + 5x = 84

⇒ 12 x = 84 ⇒ x = 7 cm

∴ Sides of the triangle are 21 cm, 28 cm, 35 cm

Also,
21^{2} + 28^{2} = 441 + 784 = 1225 = 35^{2}

⇒ The triangle is a right triangle with legs 21 cm and 28 cm.

∴ Area of the triangle = 12
× 21 × 28 cm^{2} = 294 cm2

**2. The sides of a triangle are 4 cm, 8 cm and 6 cm. Find the length of the perpendicular from the opposite vertex to the longest side.**

**Sol.**

**3. Each side of an equilateral triangle is 2 x cm. If x√3 =√48, then find its area.**

**Sol. **Area of the triangle

**4. The sides of a triangle are x, y and z. If x + y = 7 m, y + z = 9 m, and z + x = 8 m, then find the area of the triangle. **

**Sol. **We have, (x + y) + (y + z ) + (z + x) = 7 m + 9
m + 8 m

⇒ x + y + z = 12 m … (i)

Subtracting the given equations one by one from (i), we get z = 5, x = 3, y = 4

Also,
x^{2} + y^{2} = 9 + 16 = 25 = z ^{2}

⇒ The triangle is right angled with legs 3 m, 4 m.

∴
Area of the triangle = 1/2 × 3 × 4 m^{2} = 6 m^{2}.

**5. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. Find the length of its longest altitude. **

**Sol.**

∴ Area of the triangle

**6. Find the area of an isosceles triangle whose perimeter is 11cm and the base is 5 cm. **

**Sol. **Each equal side of the triangle

and semi perimeter, = 5.5 cm

**7. The area of a triangle is 30 cm2. Find the base of the altitude exceeds the base by 7 cm. **

**Sol. **Let base be x cm.

Then, altitude = (x + 7) cm.

∴ 30 = 1/2 × x × (x + 7)

⇒
x^{2} + 7x – 60 = 0 ⇒ x^{2} +
12x – 5x – 60 = 0

⇒ x (x + 12) – 5 (x + 12) = 0

⇒ (x + 12) (x – 5) = 0 ⇒ x = 12 or x = 5

⇒ x = 5 (Rejecting x = – 12)

Hence, base of the triangle = 5 cm.