Question

.The sides of a triangle are 56cm, 60cm, and 52cm long. Find the

area of the triangle.​

Answer 1

Answer:

sides of the triangle are a- 52cm, b-56cm, c- 60cm.

semi perimeter= (a+b+c)/2

=(52+56+60)/2= 84cm

by herons formula=√s(s-a)(s-b)(s-c)

=√84(84-52)(84-56)(84-60)

=√84×32×28×24

= 2×3×7×2×2×2×2×2

=1344cm sq

Answer 2

Answer:

Sides are 52cm, 56cm, and 60 cm

Area of the Triangle = ?

By Using Heron's Formula,

The area of the given triangle is;

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

Where,

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

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

Solution:

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

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

$$\begin{lgathered}\\ \implies{\sf{ \sqrt{1806336} }} \\\end{lgathered}$$

$$\begin{lgathered}\begin{lgathered}\\ \implies{\sf{ 1344 \: cm^2 }} \\\end{lgathered}\end{lgathered}$$

Hence,

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

Hope it helps!!