Math, asked by rahul5144, 6 months ago

plzzzzzz help me........ ​

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Answered by BrainlyEmpire
30

\large{\red{\bold{\underline{Given:}}}}

\sf \: Coordinates \: of \: the \: vertices \: are: \:  x(2,8), \\ \\ \: \sf \: y(-2,2) \: and \: z(6,-4).

\large{\green{\bold{\underline{To \: Find:}}}}

 \sf \: Centroid \: of \: the \: triangle

\large{\blue{\bold{\underline{Formula \: Used:}}}} \\  \\ \sf \:Coordinates \: of \:Centroid =   \: (\frac{x_{1} + x_{2} + x_{3}}{3} ),( \frac{y_{1} + y_{2} + y_{3}}{3})

\large{\red{\bold{\underline{Solution:}}}} \\  \\  \: \sf \: On \: considering \: the \: respective \: coordinates \: as :

 \sf \: x(2,8) \: \rightarrow \: (x_{1}, y_{1}) \\ \\\sf \: y(-2,2) \: \rightarrow \: (x_{2}, y_{2})  \\  \\  \sf \: z(6,-4) \: \rightarrow \: (x_{3}, y_{3})

\large{\pink{\bold{\underline{Now:}}}} \\ \\ \rightarrow \: \sf \: Centroid = ( \frac{2 + ( - 2) + 6}{3}) ,( \frac{8 + 2 + (-4)}{3} ) \\ \\  \rightarrow \sf \: Centroid = ( \frac{2 - 2 + 6}{3}) , (\frac{8 + 2 - 4}{3} ) \\  \\ \rightarrow \sf \: Centroid = ( \frac{6}{3} ),( \frac{10 - 4}{3} ) \\  \\ \rightarrow \sf \: Centroid = ( \frac{\cancel6}{\cancel3} ),( \frac{\cancel6}{\cancel3} ) \\  \\ \rightarrow \sf \: Centroid = (2,2)

\large{\orange{\bold{\underline{Therefore:}}}} \\  \\  \sf \: The \: coordinates \: of \: Centroid \: of \: the \: triangle \\ \sf \: is \: (2,2).

Answered by Anonymous
25

Answer:

\large{\red{\bold{\underline{Given:}}}}

\sf \: Coordinates \: of \: the \: vertices \: are: \:  x(2,8), \\ \\ \: \sf \: y(-2,2) \: and \: z(6,-4).

\large{\green{\bold{\underline{To \: Find:}}}}

 \sf \: Centroid \: of \: the \: triangle

\large{\blue{\bold{\underline{Formula \: Used:}}}} \\  \\ \sf \:Coordinates \: of \:Centroid =   \: (\frac{x_{1} + x_{2} + x_{3}}{3} ),( \frac{y_{1} + y_{2} + y_{3}}{3})

\large{\red{\bold{\underline{Solution:}}}} \\  \\  \: \sf \: On \: considering \: the \: respective \: coordinates \: as :

 \sf \: x(2,8) \: \rightarrow \: (x_{1}, y_{1}) \\ \\\sf \: y(-2,2) \: \rightarrow \: (x_{2}, y_{2})  \\  \\  \sf \: z(6,-4) \: \rightarrow \: (x_{3}, y_{3})

\large{\pink{\bold{\underline{Now:}}}} \\ \\ \rightarrow \: \sf \: Centroid = ( \frac{2 + ( - 2) + 6}{3}) ,( \frac{8 + 2 + (-4)}{3} ) \\ \\  \rightarrow \sf \: Centroid = ( \frac{2 - 2 + 6}{3}) , (\frac{8 + 2 - 4}{3} ) \\  \\ \rightarrow \sf \: Centroid = ( \frac{6}{3} ),( \frac{10 - 4}{3} ) \\  \\ \rightarrow \sf \: Centroid = ( \frac{\cancel6}{\cancel3} ),( \frac{\cancel6}{\cancel3} ) \\  \\ \rightarrow \sf \: Centroid = (2,2)

\large{\orange{\bold{\underline{Therefore:}}}} \\  \\  \sf \: The \: coordinates \: of \: Centroid \: of \: the \: triangle \\ \sf \: is \: (2,2).

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