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

(TRUE) Let the equal sides of the isosceles triangle be $a$ and the base be $b = 5 \, cm$.
The perimeter of the triangle is given as $11 \, cm$.
So,$a + a + 5 = 11$.
$2a = 11 - 5 = 6$.
$a = 3 \, cm$.
The area of an isosceles triangle is given by the formula $\text{Area} = \frac{b}{4} \sqrt{4a^2 - b^2}$.
Substituting the values: $\text{Area} = \frac{5}{4} \sqrt{4(3)^2 - 5^2} = \frac{5}{4} \sqrt{4(9) - 25} = \frac{5}{4} \sqrt{36 - 25} = \frac{5}{4} \sqrt{11} \, cm^2$.
Since the calculated area matches the given value,the statement is True.