Answer :

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 = 216cm^{2}

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 600cm^{2}

Now we have

Area of Quad ABCD = Area ΔABD + Area ΔBCD

Area of Quad ABCD = 216cm^{2} + 600cm^{2}

∴ Area of Quad ABCD = 816 cm^{2}.

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