Question

The sides of a triangle are$56\mathrm{cm},60\mathrm{cm}\mathrm{and}52\mathrm{cm}$ long.Then the area of the triangle is-

• A. $1322{\mathrm{cm}}^2$
• B. $1321c{m}^2$
• C. $1344{\mathrm{cm}}^2$
• D. $1392{\mathrm{cm}}^2$

Answer 1

The correct option is C

$1344{\mathrm{cm}}^2$

We know that Area of triangle$=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

According to the given details

Sides of a triangle,

$\mathrm{a}=56,\mathrm{b}=60,\mathrm{c}=52$

$\begin{array}{rcl}s& =& \frac{(\mathrm{a}+\mathrm{b}+\mathrm{c})}{2}\\ s& =& \frac{(56+60+52)}{2}\\ s& =& \frac{168}{2}\\ & =& 84.\end{array}$

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

$$\begin{gathered} =\sqrt{84(84-56)(84-60)(84-52)} \\ =\sqrt{84\times 28\times 24\times 32} \\ =1344{\mathrm{cm}}^2 \end{gathered}$$

Hence, Area of given triangle is $1344{\mathrm{cm}}^2$