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CAT Brahmagupta's Formula - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Syllabus

Quick Facts

2 Questions around this concept.

Solve by difficulty

EASY

The lengths of the sides of a cyclic quadrilateral area 1, 3, 5, and 7 units. Then the area of the quadrilateral is (in units²)

$\sqrt{121}$

$\sqrt{91}$

$\sqrt{105}$

None of these

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Concepts Covered - 1

Brahmagupta's Formula

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

$\mathrm{A=\sqrt{(s-a)(s-b)(s-c)(s-d)}}$

where s is the semiperimeter

$s=\frac{a+b+c+d}{2}$ .

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)

$\\\mathrm{Ar[CDE]=\sqrt{\left (\frac{x+y+c}{2} \right )(x+y-c)(x-y+c)(-x+y+c)}}\\\mathrm{4Ar[CDE]=\sqrt{(x+y+c)(x+y-c)(x-y+c)(-x+y+c)}}\;\;\;\;\;\ldots(1)$

But triangles ABE and CDE are similar, implying:

$\frac{Ar[A B E]}{Ar[C D E]}=\frac{a^{2}}{c^{2}}$

from which

$\\\frac{Ar[C D E]-Ar[A B E]}{Ar[C D E]}=\frac{c^2-a^{2}}{c^{2}}\\\mathrm{Let,\;A=Ar{[CDE]}-Ar[ABE]}\\\Rightarrow A=Ar[CDE]\times\left ( \frac{c^2-a^{2}}{c^{2}} \right )\;\;\;\;\;\;\;\;\;\;\;\;\ldots(2)\\\text{A is area of the Quadrilateral}$

We also have the proportions

$\\ \frac{x}{c}=\frac{y-d}{a} \\ \frac{y}{c}=\frac{x-b}{a}$

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

$x+y=c \left (\frac{b+d}{c-a} \right )$

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

$x-y=c \left (\frac{b-d}{c+a} \right )$

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

$\begin{aligned} x+y+c &=c \left (\frac{b+d}{c-a} \right )+c=c \left (\frac{b+d+c-a}{c-a} \right )=2 c \left (\frac{s-a}{c-a} \right )\\ x+y-c &=c\left ( \frac{b+d}{c-a} \right )-c=c \left (\frac{b+d-c+a}{c-a} \right )=2 c\left ( \frac{s-c}{c-a} \right ) \\ x-y+c &=c \left (\frac{b-d}{c+a} \right )+c=c\left ( \frac{b-d+c+a}{c+a} \right )=2 c \left (\frac{s-d}{c+a} \right ), \text { and } \\ -x+y+c &=-c \left (\frac{b-d}{c+a} \right )+c=c\left ( \frac{-b+d+c+a}{c+a} \right )=2 c \left (\frac{s-b}{c+a} \right ) \end{aligned}$