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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