Answer

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Hint: Here, in this problem, we will use the midpoint formula to find the midpoints of the sides of the rectangle ABCD and then will find its diagonal to see whether it's rhombus or square.

\[\begin{align}

& \text{For QR; Here, }{{x}_{1}}=2\ ;\ {{y}_{1}}\ =\ 4 \\

& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}_{2}}\ =\ 5\ ;\ {{y}_{2}}\ =\ 3\text{/}2 \\

\end{align}\]

Using distance formula

\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

\[\text{QR}\ \text{=}\ \sqrt{{{(5-2)}^{2}}\ +\ {{\left( \dfrac{3}{2}-4 \right)}^{2}}}\]

\[=\sqrt{{{(3)}^{2}}+{{\left( \dfrac{3-8}{2} \right)}^{2}}}\]

\[=\sqrt{9+{{\left( \dfrac{-5}{2} \right)}^{2}}}\]

\[=\sqrt{9+\dfrac{25}{4}}=\sqrt{\dfrac{36+25}{4}}=\sqrt{\dfrac{61}{4}}\ \text{units}\text{.}\]

\[\begin{align}

& \text{For RS}\ \text{; Here, }{{x}_{1}}=5\ ;\ {{y}_{1}}\ =\ 3\text{/}2 \\

& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}_{2}}\ =\ 2\ ;\ {{y}_{2}}\ =\ -1 \\

\end{align}\]

\[\text{RS = }\sqrt{{{(2-5)}^{2}}+{{(-1-3\text{/}2)}^{2}}}\]

\[=\sqrt{{{(-3)}^{2}}+{{\left( \dfrac{-2-3}{2} \right)}^{2}}}=\sqrt{9+{{\left( \dfrac{-5}{2} \right)}^{2}}}=\sqrt{9+\dfrac{25}{4}}\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{\dfrac{36+25}{4}}=\sqrt{\dfrac{61}{4}}\ \text{units}\text{.}\]

\[\begin{align}

& \text{For SP}\ \text{; Here, }{{x}_{1}}=2\ ;\ {{y}_{1}}\ =\ -1 \\

& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}_{2}}\ =\ -1\ ;\ {{y}_{2}}\ =\ 3\text{/}2 \\

\end{align}\]

\[\text{SP}\ \text{=}\ \sqrt{{{(-1-2)}^{2}}+{{\left( \dfrac{3}{2}+1 \right)}^{2}}}\]

\[=\sqrt{{{(-3)}^{2}}+{{\left( \dfrac{3+2}{2} \right)}^{2}}}\]

\[=\sqrt{9+\dfrac{25}{4}}\]

\[=\sqrt{\dfrac{36+25}{4}}\]

\[=\sqrt{\dfrac{61}{4}}\ \text{units}\text{.}\]

Points S is the midpoint of AD. Applying midpoint formula:

Coordinates of S:-$\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2}-,\dfrac{{{y}_{1}}+{{y}_{2}}}{2}- \right)$

Here : ${{x}_{1}}=-1;\ {{y}_{1}}=-1$

${{x}_{2}}=5;\ {{y}_{2}}=-1$

Coordinates of $S=\left( \dfrac{-1+5}{2},\dfrac{-1-1}{2} \right)=\left( \dfrac{4}{2},\dfrac{-2}{2} \right)=\left( 2,-1 \right)$

Thus,

Vertices of quadrilateral PQRS are,

$P\left( \text{-1,}{\scriptstyle{}^{3}/{}_{2}} \right),Q\left( 2,4 \right),R\left( 5,{\scriptstyle{}^{3}/{}_{2}} \right)$ and $S\left( 2,-1 \right)$

Now, finding the sides of quadrilateral PQRS.

Applying distance formula;

\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

For PQ; Here, ${{x}_{1}}=-1;\ {{y}_{1}}={\scriptstyle{}^{3}/{}_{2}}$

${{x}_{2}}=2;\ {{y}_{2}}=4$

PQ $=\sqrt{{{\left( 2-\left( -1 \right) \right)}^{2}}+{{\left( 4-{\scriptstyle{}^{3}/{}_{2}} \right)}^{2}}}$

$=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( \dfrac{8-3}{2} \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( \dfrac{5}{2} \right)}^{2}}}$

$=\sqrt{9+\dfrac{25}{4}}$

$=\sqrt{\dfrac{36+25}{4}}$

$=\sqrt{\dfrac{61}{4}}$ units.

Thus,

PQ = QR = RS = SP = $\sqrt{\dfrac{61}{4}}$ units.

$\therefore $ PQRS is a square, on a rhombus.

Now to check whether. PQRS is a square or a rhombus, we find the length of diagonals RP and SQ;

For RP:-${{x}_{1}}=-1;\ {{y}_{1}}={\scriptstyle{}^{3}/{}_{2}}$

${{x}_{2}}=5;\ {{y}_{2}}={\scriptstyle{}^{3}/{}_{2}}$

RP$=\sqrt{{{\left( 5+1 \right)}^{2}}+{{\left( \dfrac{3}{2}-\dfrac{3}{2} \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$

$\sqrt{{{6}^{2}}+0}$

6 units.

For SQ;$-{{x}_{1}}=2$ $;{{y}_{1}}=-1$

${{x}_{2}}=2$ $;{{y}_{2}}=4$

SQ$=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 4+1 \right)}^{2}}}$

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 5 \right)}^{2}}}$

$=\sqrt{{{\left( 5 \right)}^{2}}}=5$ units

Since RP $\ne $ SQ

$\therefore $ Diagonal of the quadrilateral are not equal.

So, PQRS is not a square.

Note: In this type of problems, where we are asked to find whether the given point is a rectangle that will form a rhombus or a square, we have to use the midpoint formula and the distance formula to know the length of the sides and recognize the figure formed.

__Complete step-by-step answer:__\[\begin{align}

& \text{For QR; Here, }{{x}_{1}}=2\ ;\ {{y}_{1}}\ =\ 4 \\

& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}_{2}}\ =\ 5\ ;\ {{y}_{2}}\ =\ 3\text{/}2 \\

\end{align}\]

Using distance formula

\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

\[\text{QR}\ \text{=}\ \sqrt{{{(5-2)}^{2}}\ +\ {{\left( \dfrac{3}{2}-4 \right)}^{2}}}\]

\[=\sqrt{{{(3)}^{2}}+{{\left( \dfrac{3-8}{2} \right)}^{2}}}\]

\[=\sqrt{9+{{\left( \dfrac{-5}{2} \right)}^{2}}}\]

\[=\sqrt{9+\dfrac{25}{4}}=\sqrt{\dfrac{36+25}{4}}=\sqrt{\dfrac{61}{4}}\ \text{units}\text{.}\]

\[\begin{align}

& \text{For RS}\ \text{; Here, }{{x}_{1}}=5\ ;\ {{y}_{1}}\ =\ 3\text{/}2 \\

& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}_{2}}\ =\ 2\ ;\ {{y}_{2}}\ =\ -1 \\

\end{align}\]

\[\text{RS = }\sqrt{{{(2-5)}^{2}}+{{(-1-3\text{/}2)}^{2}}}\]

\[=\sqrt{{{(-3)}^{2}}+{{\left( \dfrac{-2-3}{2} \right)}^{2}}}=\sqrt{9+{{\left( \dfrac{-5}{2} \right)}^{2}}}=\sqrt{9+\dfrac{25}{4}}\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{\dfrac{36+25}{4}}=\sqrt{\dfrac{61}{4}}\ \text{units}\text{.}\]

\[\begin{align}

& \text{For SP}\ \text{; Here, }{{x}_{1}}=2\ ;\ {{y}_{1}}\ =\ -1 \\

& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}_{2}}\ =\ -1\ ;\ {{y}_{2}}\ =\ 3\text{/}2 \\

\end{align}\]

\[\text{SP}\ \text{=}\ \sqrt{{{(-1-2)}^{2}}+{{\left( \dfrac{3}{2}+1 \right)}^{2}}}\]

\[=\sqrt{{{(-3)}^{2}}+{{\left( \dfrac{3+2}{2} \right)}^{2}}}\]

\[=\sqrt{9+\dfrac{25}{4}}\]

\[=\sqrt{\dfrac{36+25}{4}}\]

\[=\sqrt{\dfrac{61}{4}}\ \text{units}\text{.}\]

Points S is the midpoint of AD. Applying midpoint formula:

Coordinates of S:-$\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2}-,\dfrac{{{y}_{1}}+{{y}_{2}}}{2}- \right)$

Here : ${{x}_{1}}=-1;\ {{y}_{1}}=-1$

${{x}_{2}}=5;\ {{y}_{2}}=-1$

Coordinates of $S=\left( \dfrac{-1+5}{2},\dfrac{-1-1}{2} \right)=\left( \dfrac{4}{2},\dfrac{-2}{2} \right)=\left( 2,-1 \right)$

Thus,

Vertices of quadrilateral PQRS are,

$P\left( \text{-1,}{\scriptstyle{}^{3}/{}_{2}} \right),Q\left( 2,4 \right),R\left( 5,{\scriptstyle{}^{3}/{}_{2}} \right)$ and $S\left( 2,-1 \right)$

Now, finding the sides of quadrilateral PQRS.

Applying distance formula;

\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

For PQ; Here, ${{x}_{1}}=-1;\ {{y}_{1}}={\scriptstyle{}^{3}/{}_{2}}$

${{x}_{2}}=2;\ {{y}_{2}}=4$

PQ $=\sqrt{{{\left( 2-\left( -1 \right) \right)}^{2}}+{{\left( 4-{\scriptstyle{}^{3}/{}_{2}} \right)}^{2}}}$

$=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( \dfrac{8-3}{2} \right)}^{2}}}$

$=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( \dfrac{5}{2} \right)}^{2}}}$

$=\sqrt{9+\dfrac{25}{4}}$

$=\sqrt{\dfrac{36+25}{4}}$

$=\sqrt{\dfrac{61}{4}}$ units.

Thus,

PQ = QR = RS = SP = $\sqrt{\dfrac{61}{4}}$ units.

$\therefore $ PQRS is a square, on a rhombus.

Now to check whether. PQRS is a square or a rhombus, we find the length of diagonals RP and SQ;

For RP:-${{x}_{1}}=-1;\ {{y}_{1}}={\scriptstyle{}^{3}/{}_{2}}$

${{x}_{2}}=5;\ {{y}_{2}}={\scriptstyle{}^{3}/{}_{2}}$

RP$=\sqrt{{{\left( 5+1 \right)}^{2}}+{{\left( \dfrac{3}{2}-\dfrac{3}{2} \right)}^{2}}}$

$=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$

$\sqrt{{{6}^{2}}+0}$

6 units.

For SQ;$-{{x}_{1}}=2$ $;{{y}_{1}}=-1$

${{x}_{2}}=2$ $;{{y}_{2}}=4$

SQ$=\sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( 4+1 \right)}^{2}}}$

$=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 5 \right)}^{2}}}$

$=\sqrt{{{\left( 5 \right)}^{2}}}=5$ units

Since RP $\ne $ SQ

$\therefore $ Diagonal of the quadrilateral are not equal.

So, PQRS is not a square.

**$\therefore $ PQRS is Rhombus.**Note: In this type of problems, where we are asked to find whether the given point is a rectangle that will form a rhombus or a square, we have to use the midpoint formula and the distance formula to know the length of the sides and recognize the figure formed.

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