The sides of a quadrilateral ABCD are 6 cm, 8cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

Question

Philoid · Accepted Answer

We join A to C to complete the Δ ABC,

In Δ ABC,

AC² = AB² + BC²

⇒ AC² = 6² + 8²

⇒ AC² = 36 + 64

⇒ AC² = 100

⇒ AC = 10cm

Area(ΔABC) = 1/2 × AB × BC

⇒ Area(ΔABC) = 1/2 × 6 × 8

⇒ Area(ΔABC) = 24cm²

For area of ΔACD,

a = 12, b = 14, c = 10

s = (a + b + c)/2

⇒ s = (12 + 14 + 10)/2 = 36/2 = 18.

Area(ΔACD) = √s(s-a)(s-b)(s-c)

⇒ Area(ΔACD) = √18(18-12)(18-14)(18-10)

⇒ Area(ΔACD) = √18×6×4×8

⇒ Area(ΔACD) = 24√6 = 58.78cm²

Area(ABCD) = Area(ΔABC) + Area(ΔACD)

⇒ Area(ABCD) = 24 + 58.78 = 82.78 cm²