1.

Some measures are given in the figure, find the area of ABCD.

Answer»

A(ABCD) = A(∆BAD) + A(∆BDC) 

In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m 

A(∆BAD) = (1/2) x product of sides forming the right angle 

= (1/2) x l(AB) x l(AD)

= (1/2) x 40 x 9 = 180 sq. m 

In ∆BDC, l(BT) = 13m, l(CD) = 60m 

A(∆BDC) = (1/2) x base x height 

= (1/2) x l(CD) x l(BT) 

= (1/2) x 60 x 13

= 390 sq. m A (ABCD) 

= A(∆BAD) + A(∆BDC) 

= 180 + 390 

= 570 sq. m 

∴ The area of ABCD is 570 sq.m.



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