# Find the area of the cyclic quadrilateral given the two diagonals

One diagonal of a cyclic quadrilateral coincides with a diameter of a circle whose area is 36$\pi$ $cm^2$. If the other diagonal which measures 8 $cm$ meets the first diagonal at right angles, find the area of the quadrilater.

So I derived the area given for the circle and got Radius = $6$ I got stuck computing the area even though I know the two diagonals which is 12 cm and 8 cm I know that $ac+bd=d_1d_2$ I was thinking of using the $A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$ by Brahmagupta's but how? Idk the sides..

## 2 Answers

Hint: The quadrilateral is a kite.

The area of a kite is simply the product of the lengths of the two diagonals, divided by two (why?).

• $48cm^2$? is that right? I was given the brahmagupta's formula which made me think i need to use it. – Mickey Dec 21 '14 at 11:29
• @Mickey That is correct :) – Zubin Mukerjee Dec 21 '14 at 11:29

area of any quadrilateral is $\frac{1}{2}d_1(h_1+h_2)$, where $d_1$ is any diagonal and h are its perpendicular distance from other two vertices whose sum, in this case is the other diagonal.

you can derive this formula by breaking the quadrilateral into two small triangles.