Question

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

Answer 1

ar(ABCD)= ar(ABD)+ar(DBC)

$= \frac{1}{2} \times 9 \times 40 + \frac{1}{2} \times 13 \times 60 \\ 20 \times 9 + 30 \times 13 \\ 180 + 390 \\ 570 \: ans$
so area of ABCD=570m^2

Answer 2

Solution :

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Let b , h are base and corresponding

altitude of a triangle .

Area of the triangle = ( bh )/2

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Area of ABCD

= area of ∆BAD + area ∆BCD

= ( AD × AB )/2 + ( DC × BT )/2

= ( 9 × 40 )/2 + ( 60 × 13 )/2

= 9 × 20 + 30 × 13

= 180 + 390

= 570 m²

Therefore,

Area of the ∆ABCD = 570 m²

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