Question

The sum of two side of triangle is 28 cm and difference between the two is 8 cm. If the third side is third side measure 10 cm. Then find the area of triangle​

Answer 1

$\large{\mathbb{\colorbox{pink} {\boxed{\boxed{\colorbox{white} {-:Answer:-}}}}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{brown}{Given::}}}}}}$

$\pink{➠}{ \sf{a + b = 28cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i) }}$

$\pink{➠}{ \sf{a - b = 8cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(ii) }}$

$\pink{➠}{ \sf{c= 10cm }}$

$\pmb{ \bf{Focus \: point;}}$

$: : \implies \sf{a= {1}^{st} \: side \: of \: triangle}$

$: : \implies \sf{b= {2}^{nd} \: side \: of \: triangle}$

$: : \implies \sf{c= {3}^{rd} \: side \: of \: triangle}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{purple}{To \: find::}}}}}}$

$\pink{➠}{ \sf{ Area \: of \: triangle. }}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{magenta}{Formula \: used::}}}}}}$

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

$\pmb{ \bf{Where, }}$

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

$: : \implies \sf{a= {1}^{st} \: side \: of \: triangle}$

$: : \implies \sf{b= {2}^{nd} \: side \: of \: triangle}$

$: : \implies \sf{c= {3}^{rd} \: side \: of \: triangle}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{blue}{According \: to \: Question::}}}}}}$

$\pmb{ \bf{Let's \: play \: with \: numbers!!!}}$

$\pink{➠}{ \sf{a + b = 28cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{from \: equation \: (i) \} }}$

$\pink{➠}{ \sf{a - b = 8cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{from \: equation \: (ii) \} }}$

$\bf{Then,add \: equation \: (i) \: and \: (ii),we \: get}$

${: : \implies{ \sf{a + b + a - b = 28cm + 8cm }}}$

$\bf{After \: cancelling \: the \: unlike \: terms,}$

${: : \implies{ \sf{a{ \cancel {+ b}} + a { \cancel {- b }}= 28cm + 8cm }}}$

$: : \implies{ \sf{2a= 36cm }}$

$: : \implies{ \sf{a= \frac{36}{2} cm }}$

$\bf{After \: cancelling \: the \: fraction, we \: get}$

$: : \implies{ \sf{a= \cancel{\frac{36}{2}} cm}}$

$: : \implies{ \sf{a= 18cm}}$

$\bf{Then, from \: equation \: (i),we \: get}$

$\pink{➠}{ \sf{a + b = 28cm }}$

$\bf{Substituting \: the \: value \: in \: equation (i),we \: get}$

$: : \implies{ \sf{a + b = 28cm }}$

$: : \implies{ \sf{18cm + b = 28cm }}$

$: : \implies{ \sf{ b = 28cm - 18cm }}$

$: : \implies{ \sf{ b = 10cm }}$

$\pmb{ \bf{After \: that, }}$

$\bf{We \: find \: area \: of \: triangle \: with\: help \: of \: formulas;}$

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

${ \bf{Putting \: the \: values \: in \: formulas,}}$

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

${: : \implies{ \sf{ Area \: of \: triangle = \sqrt{ \frac{38cm}{2} \bigg( \frac{38cm}{2} - 18\bigg) \bigg( \frac{38cm}{2} - 10cm\bigg) \bigg( \frac{38cm}{2} -10cm \bigg) } }}}$

${ \bf{After \: cancelling \: the \: fractions,}}$

${: : \implies{ \sf{ Area \: of \: triangle = \sqrt{ \cancel{ \frac{38cm}{2}} \bigg( {\cancel \frac {38cm}{2} } - 18\bigg) \bigg( \cancel{\frac{38cm}{2}} - 10cm\bigg) \bigg( \cancel{ \frac{38cm}{2} } -10cm \bigg) } }}}$

${:: \implies{ \sf{ Area \: of \: triangle = \sqrt{19cm(19cm - 18cm)(19cm - 10cm)(19cm -10cm) } }}}$

${:: \implies{ \sf{ Area \: of \: triangle = \sqrt{19cm \times 1cm \times 9cm \times 9cm} }}}$

${:: \implies{ \sf{ Area \: of \: triangle = \sqrt{19cm \times 1cm \times { \underline{9cm \times 9cm} }} }}}$

${:: \implies{ \sf{ Area \: of \: triangle = 9\sqrt{19}cm }}}$

$\pmb{ \bf{Hence, }}$

$\pink{➠}{ \sf{ Area \: of \: triangle = 9\sqrt{19}cm }}$

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Answer 2

Step-by-step explanation:

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