Question

heron's formula chapters 12th 12.2 exercise 2nd question​

Answer 1

Step-by-step explanation:

First, construct a quadrilateral ABCD and join BD.

We know that :

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

Now, apply Pythagoras theorem in ΔBCD

BD² = BC² + CD²

⇒ BD² = 122+52

⇒ BD² = 169

⇒ BD = 13 m

Now, the area of ΔBCD = (½ ×12×5) = 30 m²

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Using Heron’s formula,

Area of ΔABD

$\sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{15(15 - 13)(15 - 9)(15 - 8)} \: m^{2} \\ = \sqrt{15 \times 2 \times 6 \times 7} \: m ^{2}$

= 6√35 m² = 35.5 m² (approximately)

∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD

= 30 m² + 35.5m² = 65.5 m²