Write True or False and justify your answer.
In a triangle,the sides are given as $11 \, cm$,$12 \, cm$,and $13 \, cm$. The length of the altitude corresponding to the side of length $12 \, cm$ is $10.25 \, cm$.

(A) Let the sides of the triangle be $a = 11 \, cm$,$b = 12 \, cm$,and $c = 13 \, cm$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{11 + 12 + 13}{2} = \frac{36}{2} = 18 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-11)(18-12)(18-13)}$
$= \sqrt{18 \times 7 \times 6 \times 5} = \sqrt{3780} = \sqrt{36 \times 105} = 6\sqrt{105} \, cm^2$.
We know that $\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}$.
Given the base is $12 \, cm$,let the altitude be $h$:
$6\sqrt{105} = \frac{1}{2} \times 12 \times h$
$6\sqrt{105} = 6h$
$h = \sqrt{105} \, cm$.
Since $\sqrt{100} = 10$ and $\sqrt{121} = 11$,$\sqrt{105} \approx 10.2469 \, cm$,which is approximately $10.25 \, cm$.
Thus,the given statement is True.