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 = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

Mid point of PQ (A) = `((-1 - 1)/2, (-1 + 4)/2)`

= `((-2)/2, 3/2)`

= `(-1, 3/2)`

Mid point of QR (B) = `((-1 + 5)/2, (4 + 4)/2) = (4/2, 8/2)` = (2, 4)

Mid point of RS (C) =`((5 + 5)/2, (4 - 1)/2) = (10/2, 3/2) = (5, 3/2)`

Mid point of PS (D) = `((5 - 1)/2, (-1 - 1)/2) = (4/2, (-2)/2)` = (2, − 1)

Distance = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

AB = `sqrt((2 + 1)^2 + (4 - 3/2)^2`

= `sqrt(3^2 + (5/2)^2`

= `sqrt(9 + 25/4)`

= `sqrt((36 + 25)/4`

= `sqrt(61/4)`

BC = `sqrt((5 - 2)^2 + (3/2 - 4)^2`

= `sqrt(9 + 25/4)`

= `sqrt((36 + 25)/4`

= `sqrt(61/4)`

CD = `sqrt((5 - 2)^2 + (3/2 + 1)^2`

= `sqrt(3^2 + (5/2)^2`

= `sqrt(9 + 25/4)`

= `sqrt((36 + 25)/4)`

= `sqrt(61/4)`

AD = `sqrt((2 + 1)^2 + (-1 - 3/2)^2`

= `sqrt(3^2 + (- 5/2)^2`

= `sqrt(9 + 25/4)`

= `sqrt(61/4)`

AB = BC = CD = AD = `sqrt(61/4)`

Since all the four sides are equal,

∴ ABCD is a rhombus.