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Question
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
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Solution
For ΔABC,
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
Therefore, ΔABC is a right-angled triangle, right-angled at point B.
Area of ΔABC`= 1/2xxABxxBC=1/2xx3xx4=6 cm^2`
For ΔADC,
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 14/2 = 7 cm
By Heron’s formula,
`"Area of triangle "=sqrt(s(s-a)(s-b)(s-c))`
`"Area of "triangle ADC=[sqrt(7(7-5)(7-5)(7-4))]cm^2`
`=(sqrt(7xx2xx2xx3))cm^2`
`=2sqrt21 cm^2`
= (2 x 4.583) cm2
= 9.166 cm2
Area of ABCD = Area of ΔABC + Area of ΔACD
= (6 + 9.166) cm2
= 15.166 cm2
= 15.2 cm2 (approximately)
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