I'd the sides of a triangle is 22m , 21m and 20m . Find the perimeter of triangle and also find the area of the triangle
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I'd the sides of a triangle is 22m , 21m and 20m . Find the perimeter of triangle and also find the area of the triangle
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⇝Appropriate Question :
The sides of a triangle is 29m , 21m and 20m . Find the perimeter of triangle and also find the area of the triangle .
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⇝Given :
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⇝To Find :
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⇝Solution :
Formula Used :
[tex]\large{\orange{\bigstar}} \: \: \: {\underline{\boxed{\red{\sf{ Perimeter{\small_{(Triangle) }} = a + b + c}}}}}[/tex]
[tex]\large{\orange{\bigstar}} \: \: \: {\underline{\boxed{\red{\sf{ Area{\small_{(Triangle) }} = \sqrt{s ( s - a) ( s - b) ( s - c)} }}}}}[/tex]
[tex]\qquad{━━━━━━━━━━━━━━━━━━━━}[/tex]
Calculating the Perimeter :
[tex]{:\longmapsto{\qquad{\sf{ Perimeter{\small_{(Triangle) }}= a + b + c}}}}[/tex]
[tex]{:\longmapsto{\qquad{\sf{ Perimeter{\small_{(Triangle) }}= 29 + 21 + 20}}}}[/tex]
[tex]{\sf{ Perimeter \: of \: Triangle \: = {\orange{\sf{ 70 \: m}}}}}[/tex]
[tex]\qquad{━━━━━━━━━━━━━━━━━━━━}[/tex]
Calculating the Area :
Semi - Perimeter :
[tex]\qquad{\implies{\sf{ S = \dfrac{a + b + c}{2}}}}[/tex]
[tex]\qquad{\implies{\sf{ S = \dfrac{29 + 21 + 20}{2}}}}[/tex]
[tex]\qquad{\implies{\sf{ S = \cancel\dfrac{70}{2}}}}[/tex]
[tex]{\sf{Semi - Perimeter \: of \: Triangle \: = {\green{\sf{ 35 \: m}}}}}[/tex]
Area :
[tex]{:\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{s (s - a) (s - b) (s - c) }}}}}[/tex]
[tex]{:\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{35 (35 - 29) (35 - 21) (35 - 20) }}}}}[/tex]
[tex]{:\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{ 35 \times 6 \times 14 \times 15}}}}}[/tex]
[tex]{\sf{ Area \: of \: Triangle \: = {\color{maroon}{\sf{ 210 \: m²}}}}}[/tex]
[tex]\qquad{━━━━━━━━━━━━━━━━━━━━}[/tex]
Therefore :
[tex]❝ {\sf{Perimeter \: of \: the \: triangle \: is \: {\blue{\underline{\pink{\sf{70 \: m}}}}}}}[/tex] .❞
[tex]❝ {\sf{Area \: of \: the \: triangle \: is \: {\blue{\underline{\pink{\sf{210 \: m²}}}}}}}[/tex] .❞
[tex]{\orange{\underline{\blue{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}}}}[/tex]
Answer:
Given :
To Find :
Solution :
[tex]\sf \underline{ \pink \bigstar \: Calculating \: the \: Perimeter : }[/tex]
[tex] \sf{:⟼ \red \: perimeter \: of \: triangle = a + b + c}[/tex]
[tex] \sf \red{:⟼ \: perimeter = 29 + 21 + 20}[/tex]
[tex] \sf \underline{ \green \bigstar \: Perimeter \: of \: triangle = \red {70 \: m}}[/tex]
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[tex] \sf \underline{ \pink \bigstar \:Calculating \: the \: Area : }[/tex]
[tex] \sf{ \hookrightarrow \: s = \frac{a + b + c}{2} }[/tex]
[tex] \sf \red{ \hookrightarrow \: s = \frac{29 + 21 + 20}{2} }[/tex]
[tex] \sf \red{ \hookrightarrow \: s = \frac{70}{2} }[/tex]
[tex] \sf \underline{ \green \bigstar \: Perimeter = \red{35 \: m}}[/tex]
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[tex] \sf \underline{ \pink \bigstar \: Area : }[/tex]
[tex] \sf { \leadsto35 (35 - 29)(35 - 1)(35 - 20)}[/tex]
[tex] \sf \red{ \leadsto35 \times 6 \times 14 \times 15}[/tex]
[tex] \sf \underline{ \green \bigstar \: Area\: of \: triangle = \red {210 {m}^{2}} }[/tex]
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Therefore :
[tex] \sf \underline{ \pink \bigstar \: Perimeter \: of \: triangle = \red {70 \: m}}[/tex]
[tex] \sf \underline{ \pink \bigstar \: Area\: of \: triangle = \red {210 {m}^{2}} }[/tex]
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