Question

The area of triangle whose sides are 42cm,34cm and 20cm in lenght is?

Answer 1

Answer :

${336 \: cm}^{2}$

$\rule{100}2$

Solution :

Sides of the given triangle :-

• 42 cm ( a )
• 34 cm ( b )
• 20 cm ( c )

First of all, find it's semi-perimeter ( s ).

We know that,

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

$\sf s =\dfrac{42+34+20}{2}$

$\sf s =\dfrac{96}{2}$

$\sf s = 48$

Also,

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

$\sf =\sqrt{48(48-42)(48-34)(48-20)}$

$</p><p>\sf =\sqrt{48(6)(14)(28)}</p><p>$

$\sf =\sqrt{2\times2\times2\times2\times3\times3\times2\times7\times2\times7\times2\times2}</p><p>$

$\sf = 2\times 2\times 3\times 2\times 7\times 2$

$\sf = 4\times 6\times 14$

$\sf = 24\times 14$

$\sf = 336\:cm^2$

Hence,

The area of given triangle is

$\bf {336 \: cm}^{2}$

$\rule{200}2$