Question

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​

Answer 1

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

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

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

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

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

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

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

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

$= \sqrt{ 16 + 16 }$

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

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

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

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

$= \frac{32}{2}$

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

Therefore.,

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

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Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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