Question

find the area of triangle whose side are the length 8 cm 11 cm and 13 cm


{ [√30 = 5.477, approx 】​

Answer 1

$\bf {\huge {\underline {\boxed {\sf\purple {Answer:\: 127 \: cm^2}}}}}$

$\huge\underline\green {Solution:}$

➡ To find Area :

♦ Semiperimeter = $\dfrac {8+11+13}{2}$

= 32

Now,

➡ Area of triangle =

$\sqrt {32-(32-8)(32-11)(32-13)}$

= $\sqrt {32 ( 24 )( 21)(19)}$

= $\sqrt {32\times 24 \times 21 \times 19}$

= $\sqrt {16128}$

= 126.99

= 127 $cm^2<strong> </strong><strong> </strong>$ ----- ( approx )

Answer :

Area of triangle is 127 $cm^2$