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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.
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Solution
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.
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