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Question
Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm ∠A = 90° and BC = CD = 52 cm.
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Solution
The figure can be drawn as follows :
Here ABD is a right triangle. So the area will be :
ΔABD = `1/2(24) (32)`
= 384
Again
BD = `sqrt(24^2 + 32^2)`
= `8sqrt(3^2 + 4^2)`
= 8 ( 5 )
= 40
Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore
DP =`1/2"BD"`
= `1/2(40)`
= 20
From the right triangle DPC we have
PC = sqrt(52^2 - 20^2)`
= `4sqrt(13^2 - 5^2)`
= 4(12)
= 48
So
ΔDPC =`1/2 (40) (48)`
= 960
Hence the area of the quadrilateral will be :
ΔABD + ΔDPC = 960 + 384
= 1344 cm2
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