Question

Find the perimeter and area of a triangle whose sides are of lenths 52cm, 56cm and 60 cm respectively

Answer 1

Step-by-step explanation:

I have post the image of the solution

Ans of area is

1560 cm

Ams of perimeter is

168 cm

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:

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

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

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

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

Hence,

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