Question

find the area of triangle two sides of which are 18 cm and 10 cm and the perimeter is 50 cm​

Answer 1

$\huge { \boxed{ \mathbb{ANSWER}}}$

$\large \underline{\frak{ \color{red}Given:}}$

$\mapsto \sf{Perimeter \: of \: triangle=50cm}$

${ \mapsto \sf{Sides \: of \: triangle=18 cm \: and \: 10cm}}$

$\large \underline{\frak{ \color{blue}To \: find:}}$

$\leadsto \sf{Area \: of \: triangle. }$

$\large \underline{\frak{ \color{b}Formula \: required:}}$

$\mp {\fbox{Perimeter \: of \: triangle=Sum of all sides = a+ b+c }} \mp$

$\mp{ \boxed{ \rm{Area \: of \: triangle= \sqrt{s( s- a)(s - b)(s - c)} }}} \mp$

$\bf{Where, }$

${ \rightarrow \sf{s=semi-perimeter \: of \: triangle.}}$

$\rightarrow \sf{a, b \: and \: c \: is \: sides \: of \: triangle. }$

$\large \underline{\frak{ \pink{Concept \: used: }}}$

$\sf{Here,the \: two \: sides \: of \: triangle \: are \: given. So, we}$

$\sf{find \: the \: third \: side \: of \: triangle. }$

$\large \underline{\frak{ \green{According \: to \: Question: }}}$

$\sf{: : \implies{Perimeter \: of \: triangle=Sum \: of \: all \: sides = a+ b+c }}$

$\sf{: : \implies{50cm=18cm + 10cm + c}}$

$\sf{: : \implies{50cm=28cm + c}}$

$\sf{: : \implies{c = 50cm - 28cm }}$

$\sf{: : \implies{c = 22cm}}$

$\bf{Hence, }$

$\longmapsto \sf{Third \: side \: of \: triangle=22cm}$

$\bf{Then, }$

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

${: : \implies{ \sf{Area \: of \: triangle= \sqrt{ \frac{50}{2} ( \frac{50}{2}cm - 18cm)( \frac{50}{2}cm - 10cm)( \frac{50}{2}cm - 22cm)} }}}$

${: : \implies{ \sf{Area \: of \: triangle= \sqrt{25cm ( 25cm - 18cm)( 25cm - 10cm)( 25cm - 22cm)} }}}$

${: : \implies{ \sf{Area \: of \: triangle= \sqrt{25cm \times 7cm \times 15cm \times 3cm} }}}$

${: : \implies{ \sf{Area \: of \: triangle= 5 \times 3\sqrt{7cm \times 5cm} }}}$

${: : \implies{ \sf{Area \: of \: triangle= 15\sqrt{35} {cm}^{2} }}}$

$\bf{Hence,}$

$\: \: \: \: \: \: \: \: \: \: \star{ \boxed{ \rm{Area \: of \: triangle=15 \sqrt{35} {cm}^{2} }}} \star$