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) = x product of sides forming the right angle

= \(\cfrac{1}{2}\)x l(AB) x l(AD) 

=\(\cfrac{1}{2}\) x 40 x 9 

= 180 sq. m 

In ∆BDC, l(BT) = 13m, l(CD) = 60m

A(∆BDC) = \(\cfrac{1}{2}\)x base x height

\(\cfrac{1}{2}\)x l(CD) x l(BT)

\(\cfrac{1}{2}\)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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