Question

Find the area of a wall which is in the shape of a parallelogram ABCD in which AB = 6 cm and BC = 11 cm and AC = 15 cm using heron's formula.
[3 MARKS]

Answer 1

Concept : 0.5 Mark
Application : 0.5 Mark
Calculation : 2 Marks

Since ABCD is a parallelogram, its opposite sides will be equal.

AB = 6 cm and BC = 11cm and AC = 15 cm

Now in triangle ABC,

Area of the triangle using heron’s formula :

$S=\frac{Perimeter}{2}=\frac{(6+11+15)}{2}=16cm$

Area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

= $\sqrt{16(16-11)(16-15)(16-6)}$

= 20 $\sqrt{2}$$c{m}^2$

Therefore the area of the parallelogram = 2(area of triangle ABC)

The area of the parallelogram = 2 × 20 $\sqrt{2}$ = 40 $\sqrt{2}$$c{m}^2$