Question

find area of traingle,two sides are 8,11 and the perimeter is 32​

Answer 1

a= 8 b=11 c=32-(8+11)=32-19=13

s= a+b+c/2= 32/2=16

$area = \sqrt{s(s - a)(s - b)(s - c)}$

a=root( 16 × (16-8)(16-11)(16-13)) =root(16×8×5×3) =root(1920)=43.817

Answer 2

✯✯ QUESTION ✯✯

Find area of traingle,two sides are 8,11 and the perimeter is 32..

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\implies\tt{Perimeter=32cm}$

$\implies\tt{Two\:sides=8cm\:and\:11cm}$

➮Firstly Lets find the third side : -

$\implies\tt{8+11+c=36}$

$\implies\tt{19+c=32}$

$\implies\tt{c=32-19}$

$\implies\tt\bold{c=13cm}$

☛So , The third side is 13cm..

➥Now ,

$\implies\tt{s=\dfrac{8+11+13}{2}}$

$\implies\tt{s=\cancel\dfrac{32}{2}=16cm}$

$\implies\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}$

$\implies\tt{\sqrt{16(16-8)(16-11)(16-13)}}$

$\implies\tt{\sqrt{16\times{8}\times{5}\times{3}}}$

$\implies\tt{\sqrt{{2}\times{2}\times{2}\times{2}\times{2}\times{}\times{2}\times{2}\times{5}\times{3}}}$

$\implies\tt{2\times{2}\times{2}\times\sqrt{2\times{5}\times{3}}}$

$\implies\tt\bold{8\sqrt{30}{cm}^{2}}$