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## 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, 212 + 282 = 441 + 784 = 1225 = 352

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

∴ Area of the triangle = 12 × 21 × 28 cm2 = 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, x2 + y2 = 9 + 16 = 25 = z 2

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

∴ Area of the triangle = 1/2 × 3 × 4 m2 = 6 m2.

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)

⇒ x2 + 7x – 60 = 0 ⇒ x2 + 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.

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