The base of an isosceles triangle measures $24 \mathrm{cm}$ and its area is $192 {\mathrm{cm}}^2$. Find its perimeter.

In the given figure, the area of the $∆ABC$ is

State ‘T’ for true and ‘F’ for false for each of the following statements:

(I) The lengths of the three sides of a triangular field are $40 \mathrm{m}, 24 \mathrm{m}$ and
$32 \mathrm{m}$ respectively. The area of the triangle $384 {\mathrm{m}}^2$.

(Il) The area of a quadrilateral $\mathrm{ABCD}$ in which $\mathrm{AB}=3 \mathrm{cm}, \mathrm{BC}=4 \mathrm{cm}, \mathrm{CD}=4 \mathrm{cm}$
$\mathrm{DA}=5 \mathrm{cm} \mathrm{and} \mathrm{AC}=5 \mathrm{cm} \mathrm{is}18 {\mathrm{cm}}^2$.

(Ill) An advertisement board is in the form of an isosceles triangle with its sides equal to $12 \mathrm{m}, 10 \mathrm{m} \mathrm{and} 10 \mathrm{m}.$ The cost of painting it at $₹2.25 \mathrm{per} {\mathrm{m}}^2$is $₹ 112.$

(IV) Heron’s formula cannot be used to calculate area of quadrilaterals.