The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively and the angle between the first two sides is a right angle. Find its area.
Options:
A .  18+24√2 cm2
B .  18+24√6 cm2
C .  36+24√2 cm2
D .  36+24√6 cm2
Answer: Option B : B Area of △ABC=12×base×height =12×6×8 =24cm2 Diagonal AC = √BC2+AB2=10cm For △ACD, s=10+12+142=18 Area =√s(s−a)(s−b)(s−c) =√18(18−10)(18−12)(18−14) =√18×8×6×4 =24√6cm2 ∴ area of quadrilateral =18+24√6cm2
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