3
$\begingroup$

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..

$\endgroup$
2
$\begingroup$

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?).

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.