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.

Asked by Pragya Singh | 1 year ago |  208

##### Solution :-

P id the mid-point of side AB,

Coordinate of P = $$\dfrac{ (-1 – 1)}{2}= \dfrac{ (-1 + 4)}{2}= -1, \dfrac{3}{2}$$

Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5,$$\dfrac{3}{2}$$)

Coordinate of S = (2, -1)

Now,

Length of PQ = $$\sqrt{(-1 – 2)}^2 + ( \dfrac{3}{2} – 4)^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}$$

Length of SP = $$\sqrt{(2+1)}^2 + (-1- \dfrac{3}{2})^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}$$

Length of QR = $$\sqrt{(2-5)}^2 + (4- \dfrac{3}{2} )^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}$$

Length of RS = $$\sqrt{(5– 2)}^2 + ( \dfrac{3}{2} +1)^2 = \dfrac{\sqrt{61}}{4}= \dfrac{\sqrt{61}}{2}$$

Length of PR (diagonal) = $$\sqrt{(-1 – 5)}^2 + ( \dfrac{3}{2} – \dfrac{3}{2} )^2 =6$$

Length of QS (diagonal) = $$\sqrt{(2– 2)}^2 + (4+1 )^2 =5$$

The above values show that, PQ = SP = QR = RS =$$\dfrac{\sqrt{61}}{2}$$, i.e. all sides are equal.

But PR ≠ QS i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.

Answered by Abhisek | 1 year ago

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