Question

Question 8
ABCD is a rectangle formed by points A (-1,-1), B(-1,4), C(5,4) and D(5,-1). P, Q, R, and S are mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS a square, rectangle or rhombus? Justify your answer.

Answer 1

Coordinates of P can be calculated as follows:
[Since, mid-point of the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)is(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$]
$(\frac{-1-1}{2},\frac{4-1}{2})=(-1,\frac{3}{2})$
Coordinates of Q can be calculated as follows:
$(\frac{5-1}{2},\frac{4+4}{2})=(2,4)$
Coordinates of R can be calculated as follows:
$(\frac{5+5}{2},\frac{4-1}{2})=(5,\frac{3}{2})$
Coordinates of S can be calculated as follows:
$(\frac{5-1}{2},\frac{-1-1}{2})=(2,-1)$
Length of PQ can be calculated as follows:
$PQ=\sqrt{{(2+1)}^2+{(4-\frac{3}{2})}^2}$
$=\sqrt{(2+1{)}^2+{\left(\frac{5}{2}\right)}^2}$
$=\sqrt{\left(9\right)+\left(\frac{25}{4}\right)}$
$=\frac{\sqrt{61}}{2}$
Similarly, QR can be calculated as follows:
$QR=\sqrt{(5-2{)}^2+{(\frac{3}{2}-4)}^2}$
$=\sqrt{3^2+{\left(\frac{-5}{2}\right)}^2}$
$=\frac{\sqrt{61}}{2}$
Similarly,
$PS=\sqrt{(2+1{)}^2+{(-1-\frac{3}{2})}^2}$
$=\sqrt{3^2+{\left(\frac{-5}{2}\right)}^2}$
$=\frac{\sqrt{61}}{2}$
Similarly, QR can be calculated as follows:
$SR=\sqrt{(2-5{)}^2+{(-1-\frac{3}{2})}^2}$
$=\sqrt{(-3{)}^2+{\left(\frac{-5}{2}\right)}^2}$
$=\frac{\sqrt{61}}{2}$
The above values shows that PQ = QR = RS = PS, i.e. all sides are equal.
Now let us calculate the diagonals.
$PR=\sqrt{(5+1{)}^2+{(\frac{3}{2}-\frac{3}{2})}^2}$
$=\sqrt{6^2}=6$
$QS=\sqrt{(2-2{)}^2+(-1-4{)}^2}$
$=\sqrt{(-5{)}^2}=5$
Hence, it is clear that while all sides are equal, diagonals are not equal. So, the given figure is a rhombus.