Q. 145.0( 1 Vote )

A field is in the

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 = 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.

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