Question

The area of the isosceles triangle is$\frac{5\sqrt{11}}{4}c{m}^2$ if the perimeter is $11cm$ and the base is $5cm$. Write True or False and justify your answer.

• A. True
• B. False

Answer 1

The correct option is A True

Step 1: Finding the sides of the triangle :

We have

Let an isosceles triangle with sides $a,b,c$

Perimeter$=11cm$

And base, $a=5$

As the triangle is isosceles,$b=c$

So, Perimeter $=a+b+c$

$\Rightarrow a+b+c=11$

$\Rightarrow 5+b+b=11$

$\Rightarrow 5+2b=11$

$$\begin{gathered} \Rightarrow 2b=6 \\ \Rightarrow b=\frac{6}{2}=3 \end{gathered}$$

So, we have,

$a=5cm,b=3cm,c=3cm$

Step 2: Finding the area :

Now, Let $s$ be the semi perimeter,

$\begin{array}{rcl}s& =& \frac{(a+b+c)}{2}\\ & =& \frac{\left(5+3+3\right)}{2}\\ & =& \frac{11}{2}\end{array}$

Now, from the Heron's Formula

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

So,

$\begin{array}{rcl}Area& =& \sqrt{\frac{11}{2}\left(\frac{11}{2}-5\right)\left(\frac{11}{2}-3\right)\left(\frac{11}{2}-3\right)}\\ & =& \sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{5}{2}\times \frac{5}{2}}\\ & =& \frac{5\sqrt{11}}{4}c{m}^2\end{array}$

Area of triangle $=\frac{5\sqrt{11}}{4}c{m}^2$

Hence, the given statement is “TRUE”