Chemistry, asked by shiva5282, 1 month ago

find the area . please don't spam

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Answers

Answered by hotcupid16
3

Answer:-

{\pmb{\frak{\underline{\underline{Given...}}}}}

\red\bigstar Sides of Triangle = 10cm, 10cm, 12cm

\begin{gathered} \end{gathered}

{\pmb{\frak{\underline{\underline{To\: Find...}}}}}

\red\bigstar Area of Triangle

\begin{gathered} \end{gathered}

{\pmb{\frak{\underline{\underline{Using \: Formulae ...}}}}}

\green\bigstar Semi Perimeter of Triangle

\dag{\underline{\boxed{\sf{\purple{Semi \:  Perimeter = \dfrac{a + b + c}{2}}}}}}

\green\bigstar Area of Triangle by herom Formula

\dag{\underline{\boxed{\sf{\purple{Area  \: of \:  Triangle = {\sqrt{s(s  -  a)(s  -  b)(s  -  c)}}}}}}}

\begin{gathered} \end{gathered}

{\pmb{\frak{\underline{\underline{Solution...}}}}}

 \ddag{\underline{\frak{\green{Firstly \:  finding \:  the  \: semi \:  perimeter \:  of  \:  Triangle..}}}}

 \quad{ : \implies{\sf{Semi \:  Perimeter =  \bf{\dfrac{a + b + c}{2}}}}}

Substituting the values

 \begin{gathered}\quad{ : \implies{\sf{Semi \:  Perimeter =  \bf{\dfrac{12  +  10  + 10}{2}}}}} \\ \\  { : \implies{\sf{Semi \:  Perimeter =  \bf{\dfrac{32}{2}}}}}  \\  \\ \qquad{ : \implies{\sf{Semi \:  Perimeter =  \bf{\cancel{\dfrac{32}{2}}}}}} \\  \\  \qquad{: \implies{\sf{Semi \:  Perimeter =  \bf{16}}}} \\  \\ \qquad{\dag{\underline{\boxed{\sf{\pink{Semi \:  Perimeter = {16}}}}}}}\end{gathered}

\begin{gathered} \end{gathered}

\ddag{\underline{\frak{\green{Now \:  finding \:  the  \: area \:  of  \:  Triangle..}}}}

 \quad{: \implies{\sf{Area  \: of \:  Triangle = \bf{\sqrt{s(s  -  a)(s  -  b)(s  -  c)}}}}}

Substituting the values

\begin{gathered}{: \implies{\sf{Area  \: of \:  \triangle = \bf{\sqrt{16(16  -  12)(16-  10)(16  -  10)}}}}} \\  \\{: \implies{\sf{Area  \: of \:   \triangle = \bf{\sqrt{16(4)(6)(6)}}}}} \\  \\   \quad\qquad{: \implies{\sf{Area  \: of \:   \triangle = \bf{\sqrt{16 \times 4 \times 6 \times 6}}}}} \\  \\ \quad{: \implies{\sf{Area  \: of \:   \triangle = \bf{\sqrt{2304}}}}} \\  \\ \qquad\qquad{: \implies{\sf{Area  \: of \:   \triangle = \bf{\sqrt{48 \times 48}}}}} \\  \\  \qquad{: \implies{\sf{Area  \: of \:   \triangle = \bf{48}}}} \\  \\  { \dag{\underline{\boxed{\sf{\pink{Area  \: of \: Triangle = \bf{48}}}}}}}\end{gathered}

Henceforth,The Area of Triangle is 48 cm.

 \begin{gathered} \end{gathered}

{\pmb{\frak{\underline{\underline{Learn \: More ...}}}}}

\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}

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Answered by jt4555
1

Answer:

The formula of the area of a triangle: 1/2 × base × height

Area= 1/2 × base × height

= 1/2 × 12 cm × 10 cm

= 1/2 × 120 cm

= 60 cm

Hence, the answer will be 60 cm

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