Question

$\large \underbrace{ \bf Question :-}$
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The perimeter of a triangular field is 240m with two sides 78m and 50m. Now, calculate the length of the altitude on the side of 50m length from its opposite vertex.
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Answer 1

Given :

• Perimeter of the Traingle = 240 m
• 1st Side of Triangle = 78 cm
• 2nd Side of Triangle = 50 m
• Altitude of Opposite Vertex = 50 m

To Find :

• Altitude of Triangle = ?

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SolutioN :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Perimeter {\small_{(Triangle)}} = a + b + c }}}}}$

• ${\underline{\boxed{\pmb{\sf{ Area {\small_{(Triangle)}} = \sqrt{ s \bigg( s - a \bigg) \bigg( s - b \bigg) \bigg( s - c \bigg) } }}}}}$

• ${\underline{\boxed{\pmb{\sf{ Area {\small_{(Triangle)}} = \dfrac{1}{2} \times Base \times Altitude }}}}}$

$\dag$ Calculating the 3rd Side :

${\longrightarrow{\qquad{\sf{ Perimeter = a + b + c }}}} \\ \\ \\ \ {\longrightarrow{\qquad{\sf{ 240 = 78 + 50 + c }}}} \\ \\ \\ \ {\longrightarrow{\qquad{\sf{ 240 = 128 + c }}}} \\ \\ \\ \ {\longrightarrow{\qquad{\sf{ 240 - 128 = c }}}} \\ \\ \\ \ {\qquad \; \; {\longrightarrow \; {\pmb{\underline{\boxed{\pink{\frak{ 3rd \; Side = 112 \; m }}}}}}}}$

$\dag$ Calculating the Area :

• Semi - Perimeter :

$\begin{gathered} \qquad \; \; \implies \; \sf { S = \dfrac{Perimeter}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; \implies \; \sf { S = \dfrac{240}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; \implies \; \sf { S = \cancel\dfrac{240}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \; \implies \; {\underline{\boxed{\pmb{\green{\frak{ Semi - Perimeter = 120 \; m }}}}}} \\ \\ \\ \end{gathered}$

• Area :

${\dashrightarrow{\qquad{\sf{ Area = \sqrt{ s \bigg( s - a \bigg) \bigg( s - b \bigg) \bigg( s - c \bigg) } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{ 120 \bigg( 120 - 78 \bigg) \bigg( 120 - 50 \bigg) \bigg( 120 - 112 \bigg) } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{ 120 \times 42 \times 70 \times 8 } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{ 2 \times 6 \times 10 \times 7 \times 6 \times 7 \times 10 \times 2 \times 4 } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 2 \times 6 \times 10 \times 7 \times \sqrt{4} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 12 \times 70 \times \sqrt{4} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 840 \times \sqrt{4} }}}} \qquad \; \; \bigg\lgroup {\purple{\sf{ Value \; of \; \sqrt{4} = 2 }}} \bigg\rgroup \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 840 \times 2 }}}} \\ \\ \\ \ {\qquad \; \; {\longrightarrow \; {\pmb{\underline{\boxed{\red{\frak{ Area = 1680 \; {m}^{2} }}}}}}}}$

$\dag$ Calculating the Altitude :

${\longmapsto{\qquad{\sf{ Area = \dfrac{1}{2} \times Base \times Altitude }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 1680 = \dfrac{1}{2} \times 50 \times Altitude }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 1680 \times 2 = 1 \times 50 \times Altitude }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 3360 = 1 \times 50 \times Altitude }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ 3360 = 50 \times Altitude }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ \dfrac{3360}{50} = Altitude }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ \cancel\dfrac{3360}{50} = Altitude }}}} \\ \\ \\ \ {\qquad \; \; {\longrightarrow \; {\pmb{\underline{\boxed{\purple{\frak{ Altitude = 67.2 \; m }}}}}}}}$

$\therefore \;$ Altitude of the Triangle is 16.2 m .

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