Math, asked by eabhilash293, 8 months ago

If (2, 1) (-2,5) are two opposite vertices of a square then then the area of the square is
0 16
0 4
O 12
0 8​

Answers

Answered by mysticd
6

 Given \: (2,1) \:and \: (-2,5) \:are \: two \: opposite \: vertices \:of \:a \: square

 Let \: (2,1) = (x_{1},y_{1})

 and \: (-2,5) = (x_{2},y_{2})

 Diagonal \: of \: the \: square (d)

 = Distance \: of \: joining \:to \: Vertices

 = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }

 = \sqrt{ (-2-2)^{2} + (5-1)^{2}}

 = \sqrt{(-4)^{2} + 4^{2} }

 = \sqrt{ 16 + 16 }

 = \sqrt{32}\: --(1)

 Now, \red{ Area \:of \: the \: square }

 = \frac{(diagonal )^{2}}{2}

 = \frac{ (\sqrt{32})^{2} }{2}

 = \frac{32}{2}

\green { = 16 \: square \:units }

Therefore.,

 \red{ Area \:of \: the \: square } \green { =16 \: square \:units }

•••♪

Answered by Anonymous
325

\huge\bf\underline\red{Question:}

\sf\gray{If \ (2, 1) \ (-2,5) \ are \ two \  opposite \ vertices \ of }

\sf\gray{a \ square \ then \  then \ the \ area \ of \ the \ square \ is \ ?}

\huge\bf\underline\pink{Solution:}

\sf\green{Given \: (2,1) \:and \: (-2,5) \:are \: two \: opposite \: vertices \:of \:a \: square}

\sf\orange{ Let \: (2,1) = (x_{1},y_{1}) }

\sf\green{ and \: (-2,5) = (x_{2},y_{2})}

\sf\orange{ Diagonal \: of \: the \: square (d) }

\sf\purple{\longrightarrow\:\:\: Distance \: of \: joining \:to \: Vertices }

\sf\blue{\longrightarrow\:\:\: \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} }}

 \sf\purple{\longrightarrow\:\:\: \sqrt{ (-2-2)^{2} + (5-1)^{2}} }

 \sf\blue{\longrightarrow\:\:\: \sqrt{(-4)^{2} + 4^{2} }}

 \sf\purple{\longrightarrow\:\:\: \sqrt{ 16 + 16 }}

 \sf\blue{\longrightarrow\:\:\:\sqrt{32}\: ........(1) }

 \sf\purple{\longrightarrow\:\:\: Area \:of \: the \: square }

 \sf\blue{\longrightarrow\:\:\: \dfrac{(Diagonal )^{2}}{2} }

 \sf\purple{\longrightarrow\:\:\:\dfrac{ (\sqrt{32})^{2} }{2} }

 \sf\blue{\longrightarrow\:\:\: \dfrac{32}{2}}

 \sf\purple{\longrightarrow\:\:\: 16 \: square \:units}

\star\:\;\sf\underline \red{ Area \:of \: the \: square \: =\:16 \: square \:units }

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