Chapter 11.8, Problem 9PSB

To determine

To find: The area of a cyclic quadrilateral with sides 1, 2, 3 and 4.

Answer to Problem 9PSB

Area of the cyclic quadrilateral is $2\sqrt{6}.$

Explanation of Solution

Given Information:

Sides of the inscribed quadrilateral are:

  $a=1,b=2,c=3\text{and}d=4$

Formula used:

Using Brahmagupta's formula,

Area of an inscribed quadrilateral $\left(A\right)=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$

Here, $s=\frac{a+b+c+d}{2}$ is the semi-perimeter of the quadrilateral

  $a,b,c\text{and}d$ are the sides of an inscribed quadrilateral

Calculation:

Let $a,b,c\text{and}d$ be the sides of the inscribed quadrilateral.

  $\therefore a=1,b=2,c=3\text{and}d=4$

We know that, Area of an inscribed quadrilateral $\left(A\right)=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$

  $"s"$ is the semi-perimeter of the quadrilateral

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

  $\begin{array}{l}s=\frac{1+2+3+4}{2}\\ s=\frac{10}{2}\\ s=5\end{array}$

So, $A=\sqrt{\left(5-1\right)\left(5-2\right)\left(5-3\right)\left(5-4\right)}$

  $\begin{array}{l}A=\sqrt{4\times 3\times 2\times 1}\\ A=\sqrt{2^2\times 3\times 2}\\ A=2\sqrt{6}\end{array}$

Hence, area of the cyclic quadrilateral is $2\sqrt{6}.$