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

P, Q, R and S are the midpoints of AB, BC, CD and DA respectively

So, coordinate of P
= ( \( \frac{-1-1}{2} , \frac{-1+4}{2} \) )
= ( \( -1 , \frac{3}{2} \) )

Similarly, coordinate of Q
= ( \( \frac{5-1}{2} , \frac{4+4}{2} \) )
= ( 2,4)

R = ( \( \frac{5+5}{2} , \frac{4-1}{2} \) ) = ( \( 5 , \frac{3}{2} \) )

S = ( \( \frac{5-1}{2} , \frac{-1-1}{2} \) ) = ( 2,-1)

Now, PQ = \( \sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \ \)
\( = \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

SP = \( \sqrt{(2+1)^2 + (-1- \frac{3}{2})^2} \ \)
\( = \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

QR = \( \sqrt{(2-5)^2 + (4- \frac{3}{2})^2} \ \)
\( = \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

RS = \( \sqrt{(5-2)^2 + ( \frac{3}{2} +1)^2} \ \)
\(= \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

PR = \( \sqrt{(-1-5)^2 + ( \frac{3}{2} - \frac{3}{2})^2 } \ \)
\( = \ 6 \)

QS = \( \sqrt{(2-2)^2 + (4+1)^2} \ \)
\( = \ 5 \)

Clearly, PQ = QR = RS = SP

But, \( PR \ne QS \)


Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.

\(\therefore \) The PQRS is a rhombus.