Question

10. Find the area of a triangle whose sides are of lengths 52cm, 56cm, and 60 cm​

Answer 1

Answer:

• Sides are 52cm, 56cm, and 60 cm
• Area of the Triangle = ?

By Using Heron's Formula,

The area of the given triangle is;

$\\ \bullet{\boxed{\sf{ Area= \sqrt{ s(s-a)(s-b)(s-c) } }}}$

Where,

$\because {\sf{\bf{ s = \dfrac{a+b+c}{2} }}}$

$\implies{\sf{ \dfrac{52+56+60}{2} }} \\ \\ \implies{\sf{ \dfrac{ \cancel{168}^{ \: \: 84}}{ \cancel{2}} }} \\ \\ \implies{\sf{ 84 \: cm}}$

Solution:

$\\ \implies{\sf{ A= \sqrt{ 84(84-52)(84-56)(84-60) } }}$

$\\ \implies{\sf{ \sqrt{ 84 \times 32 \times 28 \times 24} }}$

$\\ \implies{\sf{ \sqrt{1806336} }}$

$\\ \implies{\sf{ 1344 \: cm^2 }}$

Hence,

$\sf\pink{\bf{ The\:area\:of\: triangle\:is\:1344\:cm^2}} .$

Answer 2

Answer :– Area of ∆ = 1344 cm²