Question

8th Question
Find the area of the triangle whose size measure 52cm,56cm and 60cm respectively

Answer 1

heya user,
here is your answer,


Let three sides of the triangle be a, b, c
a = 52 cm
b = 56 cm
c = 60cm

to find the area of this triangle we will use the Heron's formula which is
$\sqrt{s(s - a)(s - b)(s - c)}$
here s indicates the semi-perimeter
s = (52+56+60)/2 = 168/2 = 84cm

area =
$\sqrt{84(84 - 52)(84 - 56)(84 - 60)} \\ = \sqrt{84 \times 32 \times 28 \times 24} \\ = \sqrt{2 \times 2 \times 3 \times 7 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 2 \times 2 \times 2 \times 3} \\ = 2 \times 3 \times 7 \times 2 \times 2 \times 2 \times 2 \times 2 \\ = 1344$

So the area of this triangle is 1344 cm²



Hope this helps

if u have any doubt or need further help ask me frankly. i would love to help.


Thank you


#Sneha

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!!