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