Chapter 11.8, Problem 7PSB

To determine

To verify: That the area of the quadrilateral given is $12\sqrt{21}+54.$

Explanation of Solution

Given Information:

Sides of the triangle are:

  $a=3,b=5\text{and}c=6$

Formula used:

Area of a triangle $\left(A\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

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

  $a,b\text{and}c$ are the sides of a triangle

Proof:

The quadrilateral has two triangles.

1^st triangle has two sides $a_1=9\text{and}{b}_1=12$

2^nd triangle has two of its sides as $a_2=10\text{and}{b}_2=11$

Let $c$ be the common side of thetwo triangles.

The first triangle is a right-angled triangle.

Using Pythagoras theorem, we have

  $c^2=a_1{}^2+b_1{}^2$

  $\begin{array}{l}{c}^2=9^2+{12}^2\\ c^2=81+144\\ c^2=225\\ c=15\end{array}$

Area of a triangle $\left(A\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

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

  $a,b\text{and}c$ are the sides of a triangle

Area of 1^st triangle $\left(A_1\right)=\sqrt{s_1\left(s_1-a_1\right)\left(s_1-b_1\right)\left(s_1-c\right)}$

Now, $s_1=\frac{a_1+b_1+c}{2}$

  $s_1=\frac{9+12+15}{2}=18$

  $\begin{array}{l}{A}_1=\sqrt{18\left(18-9\right)\left(18-12\right)\left(18-15\right)}\\ A_1=\sqrt{3^2\times 2\times 3^2\times 2\times 3\times 3}\\ A_1=3^3\times 2\\ A_1=54\end{array}$

Area of 2^nd triangle $\left(A_2\right)=\sqrt{s_2\left(s_2-a_2\right)\left(s_2-b_2\right)\left(s_2-c\right)}$

Now, $s_2=\frac{a_2+b_2+c}{2}$

  $s_2=\frac{10+11+15}{2}$

  $s_2=18$

  $\begin{array}{l}{A}_2=\sqrt{18\left(18-10\right)\left(18-11\right)\left(18-15\right)}\\ A_2=\sqrt{3^2\times 2\times 2^3\times 7\times 3}\\ A_2=12\sqrt{21}\end{array}$

Area of the quadrilateral = $A_1+A_2$

Area of the quadrilateral = $12\sqrt{21}+54$

Hence, area of the quadrilateral shown is $12\sqrt{21}+54.$