Question

Find the area of triangle whose perimeter is 42 cm and two of its sides 10 cm and 18 cm

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Perimeter of triangle :-

= 42 cm

a + b + c = 42 cm

Sides are given :-

18 + 10 + c = 42

28 + c = 42

c = 42 - 28

c = 14 cm

Hence :-

$\tt{\rightarrow s=\dfrac{42}{2}}$

s = 21 cm

${\boxed{\sf\:{Using\;Heron's\;Formula }}}$

$\tt{\rightarrow\sqrt{s(s-a)(s-b)(s-c)}}$

$\tt{\rightarrow\sqrt{21(21-18)(21-10)(21-14)}}$

$\tt{\rightarrow\sqrt{21\times 3\times 11\times 7}}$

$\tt{\rightarrow\sqrt{3\times 7\times 3\times 11\times 7}}$

$\tt{\rightarrow 21\sqrt{11}}$

Therefore we get :-

${\boxed{\sf\:{Area\;of\;triangle\;is\;21\sqrt{11}}}}$

Answer 2

${\textbf{Answer}}$