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Question
Find the area of a quadrilateral ABCD is which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution
For ΔPQR
`PQ^2=QR^2+RP^2`
`(5^2)=(3)^2+(4)^2` [∵𝑃𝑅=3 𝑄𝑅=4 𝑎𝑛𝑑 𝑃𝑄=5]
So, ΔPQR is a right angled triangle. Right angle at point R.
Area of ΔABC =`1/2xxQRxxRP`
`=1/2xx3xx4`
`=6cm^2`
For ΔQPS
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5)cm = 14 cm
S = 7 cm
By Heron’s formulae
Area of Δle`sqrt(s(s-a)(s-b)(s-c))cm^2`
Area of Δle PQS `=sqrt(7(7-5)(7-4)(7-3))cm^2`
`=2sqrt(21)cm^2`
`=(2xx4.583)cm^2`
`=9.166cm^2`
Area of PQRS = Area of PQR + Area of ΔPQS = `(6+9.166)cm^2=15.166 cm^2`
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