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.

Answer 1

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`