Question

Sides of triangle are 3cm, 4cm and 5cm. Find area of triangle by heron's FORMULA​

Answer 1

Diagram:-

$\setlength{\unitlength}{0.8cm} \begin{picture}(6,5) \thicklines \put(1,0.5){\line(2,1){3}} \put(4,2){\line(-2,1){2}} \put(2,3){\line(-2,-5){1}} \put(0.7,0.3){ A } \put(4.05,1.9){ B } \put(1.7,2.95){ C } \put(3.1,2.5){ a } \put(1.3,1.7){ b } \put(2.5,1.05){ c } \put(0.3,4){ F=\sqrt{s(s-a)(s-b)(s-c)} } \put(3.5,0.4){ \displaystyle s:=\frac{a+b+c}{2} } \end{picture}$

Given:

• Sides of triangle:- a = 3cm, b = 4cm and c = 5cm

Find:

• Area of triangle by heron's formula

Solution:

we, know that

$\boxed{ \to \red{\sf semi - perimeter = \dfrac{a + b + c}{2} }}$

where,

• a = 3cm
• b = 4cm
• c = 5cm

So,

$\to\sf \purple{s= \dfrac{a + b + c}{2} }$

$\to\sf \green{ s= \dfrac{3 + 4 + 5}{2} }$

$\to\sf \pink{ s= \cancel{\dfrac{12}{2} } = 6cm}$

$\to\sf \red{ s = 6cm}$

So, s = 6cm

$\rule{300}{3}$

We, know that

$\boxed{ \to \green{\sf Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} }}$

where,

• s = 6cm
• a = 3cm
• a = 3cmb = 4cm
• a = 3cmb = 4cmc = 5cm

So,

$\to\sf \blue{ Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} }$

$\to\sf \pink{Area \: of \: triangle = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} }$

$\to\sf \purple{ Area \: of \: triangle = \sqrt{6(3)(2)(1)} }$

$\to\sf \orange{ Area \: of \: triangle = \sqrt{6(6)(1)} }$

$\to\sf \green{ Area \: of \: triangle = \sqrt{6 \times 6} }$

$\to\sf \purple{ Area \: of \: triangle = \sqrt{36} {cm}^{2} }$

$\to\sf \red{ Area \: of \: triangle = 6 {cm}^{2} }$

$\rule{300}{3}$

Hence, area of Triangle = 6cm²

Answer 2

Step-by-step explanation:

hope this helps you

please mark this answer as brainalist and follow me fast to get best answer