Question

\[{\text{ABCD}}\] is a rectangle formed by the points, \[{\text{A}}\left( { - 1, - 1} \right),{\text{ B}}\left( { - 1,4} \right),{\text{ C}}\left( {5,4} \right)\] and \[{\text{D}}\left( {5, - 1} \right).\] \[{\text{P, Q, R, S}}\] are the mid-points of \[{\text{AB,BC,CD }}\] and \[{\text{DA}}\] respectively. Is the quadrilateral PQRS a square? A rectangle? Or a rhombus? Justify your answer.

Answer 1

Hint: Since \[{\text{P,Q,R,S}}\] are the mid-points of side \[{\text{AB, BC, CD }}\] and \[{\text{DA}}\] of rectangle \[{\text{ABCD}}\], the question can be solved by applying the midpoint formula through which we will get the coordinates of \[{\text{P,Q,R,S}}\] and finally we have to apply the distance formula and we will get a length of each side.

Complete step-by-step answer:
               

Firstly, applying midpoint formula, i.e., \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]
Here \[{\text{P}}\] is the midpoint of \[{\text{AB}}\], where \[{x_1} = - 1,{y_1} = - 1\] and \[{x_2} = - 1,{y_2} = 4\]
We get, Coordinates of \[{\text{P}} = \left( {\dfrac{{ - 1 - 1}}{2}, \dfrac{{ - 1 + 4}}{2}} \right) = \left( { - \dfrac{2}{2},\dfrac{3}{2}} \right) = \left( { - 1,\dfrac{3}{2}} \right)\]
Here \[{\text{Q}}\] is the midpoint of \[{\text{BC}}\], where \[{x_1} = - 1,{y_1} = 4\] and \[{x_2} = 5,{y_2} = 4\]
We get, Coordinates of \[{\text{Q = }}\left( {\dfrac{{ - 1 + 5}}{2}, \dfrac{{4 + 4}}{2}} \right) = \left( {\dfrac{4}{2}, \dfrac{8}{2}} \right) = \left( {2,4} \right)\]
Here \[{\text{R}}\] is the midpoint of \[{\text{CD}}\], where \[{x_1} = 5,{y_1} = 4\]and \[{x_2} = 5,{y_2} = - 1\]
Coordinates of ${\text{R}} = \left( {\dfrac{{5 + 5}}{2}, \dfrac{{ - 1 + 4}}{2}} \right) = \left( {\dfrac{{10}}{2},\dfrac{3}{2}} \right) = \left( {5, \dfrac{3}{2}} \right)$
Here \[{\text{S}}\] is the midpoint of \[{\text{DA}}\], where \[{x_1} = 5,{y_1} = - 1\] and \[{x_2} = - 1,{y_2} = - 1\]
Coordinates of $S = \left( {\dfrac{{ - 1 + 5}}{2}, \dfrac{{ - 1 - 1}}{2}} \right) = \left( {\dfrac{4}{2}, - \dfrac{2}{2}} \right) = \left( {2, - 1} \right)$,
Therefore, vertices of the quadrilateral PQRS are $P\left( { - 1, \dfrac{3}{2}} \right)$, $Q(2,4)$, $R\left( {5,\dfrac{3}{2}} \right)$, $S(2,1)$
               

Now, by applying distance formula, i.e., $\sqrt {({x_2} - {x_1})_{}^2 + ({y_2} - {y_1})_{}^2} $, we can find out the length of all the sides of the quadrilateral \[{\text{PQRS}}\]
So, length of side $PQ = \sqrt {( - 1 - 2)_{}^2 + \left( {\dfrac{3}{2} - 4} \right)_{}^2} = \sqrt {( - 3)_{}^2 + \left( { - \dfrac{5}{2}} \right)_{}^2} = \sqrt {9 + \dfrac{{25}}{4}} = \sqrt {\dfrac{{61}}{4}} $
Here ${x_2} = - 1,{x_1} = 2,{y_2} = \dfrac{3}{2},{y_1} = 4$
Length of side $QR = \sqrt {(2 - 5)_{}^2 + \left( {4 - \dfrac{3}{2}} \right)_{}^2} = \sqrt {( - 3)_{}^2 + \left( {\dfrac{5}{2}} \right)_{}^2} = \sqrt {9 + \dfrac{{25}}{4}} = \sqrt {\dfrac{{61}}{4}} $
Here ${x_2} = 2, {x_1} = 5, {y_2} = - 3, {y_1} = \dfrac{5}{2}$
Length of side $RS = \sqrt {(5 - 2)_{}^2 + \left( {\dfrac{3}{2} + 1} \right)_{}^2} = \sqrt {(3)_{}^2 + \left( {\dfrac{5}{2}} \right)_{}^2} = \sqrt {9 + \dfrac{{24}}{4}} = \sqrt {\dfrac{{61}}{4}} $
Here ${x_2} = 5, {x_1} = 2, {y_2} = \dfrac{3}{2}, {y_1} = - 1$
Length of side $SP = \sqrt {(2 + 1)_{}^2 + \left( { - 1 - \dfrac{3}{2}} \right)_{}^2} = \sqrt {(3)_{}^2 + \left( { - \dfrac{5}{2}} \right)_{}^2} = \sqrt {9 + \dfrac{{25}}{4}} = \sqrt {\dfrac{{61}}{4}} $
Here ${x_2} = 2, {x_1} = - 1, {y_2} = - 1, {y_1} = \dfrac{3}{2}$
Since all the sides of the quadrilateral \[{\text{PQRS}}\] so it is clear that it can be either square or rhombus. So now we have to find out the length of the diagonal to check whether it is a square or quadrilateral.
Here, we applying distance formula $\sqrt {({x_2} - {x_1})_{}^2 + ({y_2} - {y_1})_{}^2} $
We get, Length of the diagonal $PR = \sqrt {( - 1 - 5)_{}^2 + \left( {\dfrac{3}{2} - \dfrac{3}{2}} \right)_{}^2} = \sqrt {( - 6)_{}^2 + 0 = } \sqrt {36} = 6$ ,
Here ${x_2} = - 1, {x_1} = 5, {y_2} = \dfrac{3}{2}, {y_1} = \dfrac{3}{2}$
Length of the diagonal $QS = \sqrt {(2 - 2)_{}^2 + (4 + 1)_{}^2} = \sqrt {0 + (5)_{}^2} = \sqrt {25} = 5$
Here ${x_2} = 2, {x_1} = 2, {y_2} = 4, {y_1} = - 1$
Since the length of the diagonals of the quadrilateral \[{\text{PQRS}}\] are not equal but the length of the sides are equal so it is a rhombus.

Note: There are six types of quadrilateral like rhombus, square, parallelogram, rectangle, trapezium and cyclic quadrilateral.
In the case of squares, all the sides and diagonals are of equal length.
In the case of rhombus all sides are equal, and the diagonals are not equal and form right angles.