Chapter 11.8, Problem 6PSB

a.

To determine

To find: The area of a triangle with sides 3, 5 and6.

Answer to Problem 6PSB

The area of a triangle is $2\sqrt{2}.$

Explanation of Solution

Given Information:

Sides of the triangle are:

  $a=3,b=5\text{and}c=6$

Formula used:

Using Heron’s formula,

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

Calculation:

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

  $\therefore a=3,b=5\text{and}c=6$

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

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

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

  $\begin{array}{l}s=\frac{3+5+6}{2}\\ s=\frac{14}{2}\\ s=7\end{array}$

So, $A=\sqrt{7\left(7-3\right)\left(7-5\right)\left(7-6\right)}$

  $\begin{array}{l}A=\sqrt{7\times 4\times 2\times 1}\\ A=2\sqrt{14}\end{array}$

So, the area of a triangle is $2\sqrt{2}.$

Hence, area of the triangle with sides 3, 5 and 6 is $2\sqrt{2}.$

b.

To determine

To calculate: The area of a triangle with sides 7, 20 and 23.

Answer to Problem 6PSB

The area of the triangle is $30\sqrt{5}.$

Explanation of Solution

Given Information:

Sides of the triangle are:

  $a=7,b=20\text{and}c=23$

Formula used:

Using Heron’s formula,

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

Calculation:

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

  $\therefore a=7,b=20\text{and}c=23$

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

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

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

  $\begin{array}{l}s=\frac{7+20+23}{2}\\ s=\frac{50}{2}\\ s=25\end{array}$

So, $A=\sqrt{25\left(25-7\right)\left(25-20\right)\left(25-23\right)}$

  $\begin{array}{l}A=\sqrt{5^2\times 18\times 5\times 2}\\ A=\sqrt{5^3\times 3^2\times 2\times 2}\\ A=30\sqrt{5}\end{array}$

Hence, the area of the triangle with sides 7, 20 and 23 is $30\sqrt{5}.$