Question

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.

Answer 1

Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm.

Now, join AC.

We have, ABC is a right-angled triangled at B.

Now, AC² = AB² + BC²  ...[By Pythagoras theorem]

= 6² + 8²

= 36 + 64

= 100

⇒ AC = 10 cm  ...[Taking positive square root]

∴ Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD  ...(i)

Now, area of ΔABC = `1/2 xx AB xx BC`  ...[∵ Area of triangle = `1/2` (base × height)]

= `1/2 xx 6 xx 8`

= 24 cm²

In ΔACD, AC = a = 10 cm, CD = b = 12 cm

And DA = c = 14 cm

Now, semi-perimeter of ΔACD,

`s = (a + b + c)/2`

= `(10 + 12 + 14)/2`

= `36/2`

= 18 cm

Area of ΔACD = `sqrt(s(s - a)(s - b)(s - c))`  ...[By Heron’s formula]

= `sqrt(18(18 - 10)(18 - 12)(18 - 14))`

= `sqrt(18 xx 8 xx 6 xx 4)`

= `sqrt((3)^2 xx 2 xx 4 xx 2 xx 3 xx 2 xx 4)`

= `3 xx 4 xx 2 sqrt(3 xx 2)`

= `24sqrt(6)  cm^2`

From equation (i),

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

= `24 + 24sqrt(6)`

= `24(1 + sqrt(6))  cm^2`

Hence, the area of quadrilateral is `24(1 + sqrt(6))  cm^2`.