Chapter 11.8, Problem 3PSA

a.

To determine

To find: The area of an inscribed quadrilateral with sides 5, 7, 4 and 10.

Answer to Problem 3PSA

Area of the inscribed quadrilateral is $36{\text{ units}}^2.$

Explanation of Solution

Given Information:

Sides of the inscribed quadrilateral are:

  $a=5,b=7,c=4\text{and}d=10$

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.

  $\therefore a=5,b=7,c=4\text{and}d=10$

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

  $\begin{array}{l}s=\frac{5+7+4+10}{2}\\ s=\frac{26}{2}\\ s=13\end{array}$

So, $A=\sqrt{\left(13-5\right)\left(13-7\right)\left(13-4\right)\left(13-10\right)}$

  $\begin{array}{l}A=\sqrt{8\times 6\times 9\times 3}\\ A=\sqrt{2^3\times 2\times 3\times 3^2\times 3}\\ A=\sqrt{2^4\times 3^4}\\ A=2^2\times 3^2\\ A=36\end{array}$

Hence, area of the inscribed quadrilateral is $36{\text{ units}}^2$ .

b.

To determine

To calculate: The area of an inscribed quadrilateral with sides 2, 4, 5 and 9.

Answer to Problem 3PSA

Area of the inscribed quadrilateral is $4\sqrt{15}$

  ${\text{units}}^2$

Explanation of Solution

Given Information:

Sides of the inscribed quadrilateral are:

  $a=2,b=4,c=5\text{and}d=9$

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.

  $\therefore a=2,b=4,c=5\text{and}d=9$

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

  $\begin{array}{l}s=\frac{2+4+5+9}{2}\\ s=\frac{20}{2}\\ s=10\end{array}$

So, $A=\sqrt{\left(10-2\right)\left(10-4\right)\left(10-5\right)\left(10-9\right)}$

  $\begin{array}{l}A=\sqrt{8\times 6\times 5\times 1}\\ A=\sqrt{2^3\times 2\times 3\times 5}\\ A=\sqrt{2^4\times 3\times 5}\\ A=2^2\sqrt{3\times 5}\\ A=4\sqrt{15}\end{array}$

Hence, area of the inscribed quadrilateral is $4\sqrt{15}{\text{units}}^2.$

c.

To determine

To find: Area of an inscribed quadrilateral with sides 3, 5, 9 and 5.

Answer to Problem 3PSA

Area of the inscribed quadrilateral with sides 3, 5, 9 and 5 is ${\text{24 units}}^2.$

Explanation of Solution

Given Information:

Sides of the inscribed quadrilateral are:

  $a=3,b=5,c=9\text{and}d=5$

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.

  $a=3,b=5,c=9\text{and}d=5$

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

  $\begin{array}{l}s=\frac{3+5+9+5}{2}\\ s=\frac{22}{2}\\ s=11\end{array}$

So, $A=\sqrt{\left(11-3\right)\left(11-5\right)\left(11-9\right)\left(11-5\right)}$

  $\begin{array}{l}A=\sqrt{8\times 6\times 2\times 6}\\ A=\sqrt{2^3\times 2\times 3\times 2\times 2\times 3}\\ A=\sqrt{2^6\times 3^2}\\ A=2^3\times 3\\ A=24\end{array}$

Hence, area of the inscribed quadrilateral is $24{\text{units}}^2.$

d.

To determine

To calculate: The area of an inscribed quadrilateral with sides 1, 5, 9 and 11.

Answer to Problem 3PSA

Area of the inscribed quadrilateral is $16\sqrt{3}{\text{units}}^2.$

Explanation of Solution

Given Information:

Sides of the inscribed quadrilateral are:

  $a=1,b=5,c=9\text{and}d=11$

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.

  $\therefore a=1,b=5,c=9\text{and}d=11$

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

  $\begin{array}{l}s=\frac{1+5+9+11}{2}\\ s=\frac{26}{2}\\ s=13\end{array}$

So, $A=\sqrt{\left(13-1\right)\left(13-5\right)\left(13-9\right)\left(13-11\right)}$

  $\begin{array}{l}A=\sqrt{12\times 8\times 4\times 2}\\ A=\sqrt{2^2\times 3\times 2^3\times 2^2\times 2}\\ A=\sqrt{2^8\times 3}\\ A=2^4\sqrt{3}\\ A=16\sqrt{3}\end{array}$

Hence, area of the inscribed quadrilateral is $16\sqrt{3}{\text{ units}}^2$ .