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Question
Find the area of the quadrilateral ABCD in which in AB=42 cm, BC=21 cm, CD=29cm, DA=34 cm and diagonal BD = 20 cm.
Solution
Area of ΔABD=`sqrt(s(s-a)(s-b)(s-c))`
`s=1/2(a+b+c)`
`s=(42+20+34)/2`
`s= 48 cm`
Area of ` ΔABD=sqrt(48(48-42)(48-20)(48-34))`
=`sqrt(48xx6xx28xx14)`
=`sqrt(112896)`
=`336 cm^2`
Area of` Δ BDC=sqrt(s(s-a)(s-b)(s-c))`
s=`1/2 (a+b+c)`
s=`(21+20+29)/2`
s=`35 cm`
Area of Δ BDC= `sqrt(35(35-29)(35-20)(35-21))`
=`sqrt(35xx6xx15xx14)`
= `sqrt(44100)`
=`210cm^2`
∴ Area of quadrilateral ABCD = Area of ΔABD+ Area of ΔBDC
=`336+210`
=`546cm^2`
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