[tex] \\ \\ [/tex]
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
[tex] \dag [/tex] Formula Used :
Where :
[tex] \dag [/tex] Calculating the Semi - Perimeter :
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ a + b + c }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ 10 + 10 + 12 }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ 32 }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \cancel\dfrac{ 32 }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Semi - Perimeter = 16 \; cm }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \\ \end{gathered} [/tex]
[tex] \dag [/tex] Calculating the Area :
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ \bigg( s - a \bigg) \bigg( s - b \bigg) \bigg( s - c \bigg) } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ 16 \bigg( 16 - 10 \bigg) \bigg( 16 - 10 \bigg) \bigg( 16 - 12 \bigg) } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ 16 \times 6 \times 6 \times 4 } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ \underline{4 \times 4} \times \underline{6 \times 6} \times 4 } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 4 \times 6 \times \sqrt{4} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 24 \times \sqrt{4} } \qquad \; \; \bigg\lgroup {\pink{\sf{ Value \; of \; \sqrt{4} = 2 }}} \bigg\rgroup \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 24 \times 2 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Area = 48 \; {cm}^{2} }}}}} \; {\purple{\pmb{\bigstar}}} \\ \\ \\ \end{gathered} [/tex]
[tex] \therefore \; [/tex] Area of the Triangle is 48 cm² .
:)
[tex] \small \dag \sf \blue{ \underline{ \red{question}}}[/tex]
find the area of an isosceles triangle whose sides are 10 cm and base is 12cm.
[tex] \small \dag \sf \blue{ \underline{ \red{Answer :-}}}[/tex]
➻[tex]\sf\purple{\underline{FORMULA}}[/tex]
[tex] \sqrt{s(s - a)(s - b)(s - c)}[/tex]
➻[tex]\sf\purple{\underline{SEMI PERIMETER = A+B+C/2}}[/tex]
⟶ 10+10+12/2
⟶ 32/2
⟶ 16
[tex]\underline{\rule{220pt}{3pt}}[/tex]
➻[tex]\large\bigstar\mathtt{AREA}[/tex]
(s - a) = 16 - 10 = 6
(s - b) = 16 - 10 = 6
(s - c) = 16 - 12 = 4
[tex] \sqrt{16 \: \times 6 \times 6 \times 4} \\ 2\sqrt{4 \times 4 \times 6 \times 6} \\ 2 \times 4 \times 6 \\8 \times 6 \\ 48[/tex]
[tex]\sf\mathfrak \purple{\underline{Final\:Answer:-}}[/tex]
Area of the Triangle = [tex]\fcolorbox{red}{blue}{ 48 cm } [/tex]
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Hope it helps you
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Answers & Comments
Given :
[tex] \\ \\ [/tex]
To Find :
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
SolutioN :
[tex] \dag [/tex] Formula Used :
Where :
[tex] \\ \\ [/tex]
[tex] \dag [/tex] Calculating the Semi - Perimeter :
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ a + b + c }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ 10 + 10 + 12 }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ 32 }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \cancel\dfrac{ 32 }{2} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Semi - Perimeter = 16 \; cm }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \\ \end{gathered} [/tex]
[tex] \\ \\ [/tex]
[tex] \dag [/tex] Calculating the Area :
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ \bigg( s - a \bigg) \bigg( s - b \bigg) \bigg( s - c \bigg) } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ 16 \bigg( 16 - 10 \bigg) \bigg( 16 - 10 \bigg) \bigg( 16 - 12 \bigg) } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ 16 \times 6 \times 6 \times 4 } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ \underline{4 \times 4} \times \underline{6 \times 6} \times 4 } } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 4 \times 6 \times \sqrt{4} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 24 \times \sqrt{4} } \qquad \; \; \bigg\lgroup {\pink{\sf{ Value \; of \; \sqrt{4} = 2 }}} \bigg\rgroup \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 24 \times 2 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Area = 48 \; {cm}^{2} }}}}} \; {\purple{\pmb{\bigstar}}} \\ \\ \\ \end{gathered} [/tex]
[tex] \\ \\ [/tex]
[tex] \therefore \; [/tex] Area of the Triangle is 48 cm² .
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
:)
Verified answer
[tex] \small \dag \sf \blue{ \underline{ \red{question}}}[/tex]
find the area of an isosceles triangle whose sides are 10 cm and base is 12cm.
[tex] \small \dag \sf \blue{ \underline{ \red{Answer :-}}}[/tex]
GIVEN :
➻[tex]\sf\purple{\underline{FORMULA}}[/tex]
[tex] \sqrt{s(s - a)(s - b)(s - c)}[/tex]
➻[tex]\sf\purple{\underline{SEMI PERIMETER = A+B+C/2}}[/tex]
⟶ 10+10+12/2
⟶ 32/2
⟶ 16
[tex]\underline{\rule{220pt}{3pt}}[/tex]
➻[tex]\large\bigstar\mathtt{AREA}[/tex]
(s - a) = 16 - 10 = 6
(s - b) = 16 - 10 = 6
(s - c) = 16 - 12 = 4
[tex] \sqrt{16 \: \times 6 \times 6 \times 4} \\ 2\sqrt{4 \times 4 \times 6 \times 6} \\ 2 \times 4 \times 6 \\8 \times 6 \\ 48[/tex]
[tex]\sf\mathfrak \purple{\underline{Final\:Answer:-}}[/tex]
Area of the Triangle = [tex]\fcolorbox{red}{blue}{ 48 cm } [/tex]
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Hope it helps you