Question

PQRS is a rectangle formed by joining the points P(– 1, – 1), Q(– 1, 4), R(5, 4) and S(5, – 1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.

Answer 1

Mid point of a line = ${\displaystyle (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})}$

Mid point of PQ (A) = ${\displaystyle (\frac{-1-1}{2},\frac{-1+4}{2})}$

= ${\displaystyle (\frac{-2}{2},\frac{3}{2})}$

= ${\displaystyle (-1,\frac{3}{2})}$

Mid point of QR (B) = ${\displaystyle (\frac{-1+5}{2},\frac{4+4}{2})=(\frac{4}{2},\frac{8}{2})}$ = (2, 4)

Mid point of RS (C) =${\displaystyle (\frac{5+5}{2},\frac{4-1}{2})=(\frac{10}{2},\frac{3}{2})=(5,\frac{3}{2})}$

Mid point of PS (D) = ${\displaystyle (\frac{5-1}{2},\frac{-1-1}{2})=(\frac{4}{2},\frac{-2}{2})}$ = (2, − 1)

Distance = ${\displaystyle \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$

AB = ${\displaystyle \sqrt{{(2+1)}^2+{(4-\frac{3}{2})}^2}}$

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

= ${\displaystyle \sqrt{9+\frac{25}{4}}}$

= ${\displaystyle \sqrt{\frac{36+25}{4}}}$

= ${\displaystyle \sqrt{\frac{61}{4}}}$

BC = ${\displaystyle \sqrt{{(5-2)}^2+{(\frac{3}{2}-4)}^2}}$

= ${\displaystyle \sqrt{9+\frac{25}{4}}}$

= ${\displaystyle \sqrt{\frac{36+25}{4}}}$

= ${\displaystyle \sqrt{\frac{61}{4}}}$

CD = ${\displaystyle \sqrt{{(5-2)}^2+{(\frac{3}{2}+1)}^2}}$

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

= ${\displaystyle \sqrt{9+\frac{25}{4}}}$

= ${\displaystyle \sqrt{\frac{36+25}{4}}}$

= ${\displaystyle \sqrt{\frac{61}{4}}}$

AD = ${\displaystyle \sqrt{{(2+1)}^2+{(-1-\frac{3}{2})}^2}}$

= ${\displaystyle \sqrt{3^2+{(-\frac{5}{2})}^2}}$

= ${\displaystyle \sqrt{9+\frac{25}{4}}}$

= ${\displaystyle \sqrt{\frac{61}{4}}}$

AB = BC = CD = AD = ${\displaystyle \sqrt{\frac{61}{4}}}$

Since all the four sides are equal,

∴ ABCD is a rhombus.