CAT Brahmagupta's Formula - Practice Questions & MCQ

Brahmagupta's formula provides the area A of a cyclic quadrilateral (i.e., a simple quadrilateral that is inscribed in a circle) with sides of length a, b, c, and d as

where s is the semiperimeter

.

Proof:

Let the quadrilateral be ABCD, with AB = a, BC = b, CD = c and AD = d. Extend AD and BC to meet at E, outside the circumcircle:

(If AD∥BC then considers the other pair of the opposite sides. If those two are also parallel, the quadrilateral is a rectangle, and Brahmagupta's formula reduces to the standard formula for the area of a rectangle.)

Let CE = x and DE = y.

Then area of ∆CDE (using Heron's Formula)

But triangles ABE and CDE are similar, implying:

from which

We also have the proportions

Adding the two and solving for (x+y) gives

Similarly, subtracting one from the other and solving for x−y we obtain

from which we can find all the terms in Heron's formula. For example,

Substituting the values in (1), we get

Using (2) and (3), we get