A field is in the shape of a quadrilateral ABCD in which side AB = 18m side AD = 24m side BC = 40m DC = 50m and ∠A = 90°. Find the area of the field.

The figure is given below:

In the figure we have ∠A = 90°

So we have in ΔDAB

By pythogares theorem

(BD)² = (AB)² + (AD)²

⇒ (BD)² = (18)² + (24)²

⇒ (BD)² = 324 + 576 = 900

⇒ BD = = 30cm.

Area of Quad ABCD = Area ΔABD + Area ΔBCD

∴ Area ΔABD =

⇒

∴ Area ΔABD = 216cm²

Now for area of ΔBCD

We have BC = 40cm, CD = 50cm, BD = 30cm.

We can assume BC = b = 40cm and CD = c = 50cm and BD = d = 30cm.

∴ By Heron’s formula we get area of triangle

=

(where s = s is called as the semi perimeter )…(1)

S =

∴ Substituting the value of s, a, b and c in equation 1 we get

⇒

=

=

∴ The area of ΔBCD is 600cm²

Now we have

Area of Quad ABCD = Area ΔABD + Area ΔBCD

Area of Quad ABCD = 216cm² + 600cm²

∴ Area of Quad ABCD = 816 cm².