1.ABCD is a rectangle ans P, Q, R and S are mid point of the sides AB, BC, CD, and SA respectively. Show that the quadrilateral PQRS is Rhombus.
Answers
Answer:
Let us join AC and BD
In AABC,
P and Q are the mid-points of AB and BC
respectively.
:. PQ || AC and PQ = AC (Mid-point theorem) ...
(1)
Similarly in AADC,
SR || AC and SR = AC (Mid-point theorem) ... (2)
Clearly, PQ || SR and PQ = SR
=
Since in quadrilateral PQRS, one pair of
opposite sides is equal and parallel to
each other, it is a parallelogram.
.. PS II QR and PS = QR (Opposite sides of
parallelogram)... (3)
In ABCD, Q and Rare the mid-points of side BC
and CD respectively.
:. QR || BD and QR =BD (Mid-point theorem) ...
(4)
However, the diagonals of a rectangle are
equal.
.:. AC = BD ...(5)
=
By using equation (1), (2), (3), (4), and (5), we
obtain
PQ = QR = SR = PS
=
Therefore, PQRS is a rhombus
Step-by-step explanation:
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Let us join AC and BD. In △ABC,
P and Q are the mid-points of AB and BC respectively.
∴ PQ || AC and PQ = 1/2 AC (Mid-point theorem) ... (1)
Similarly, in △ADC,
SR || AC and SR = 1/2 AC (Mid-point theorem) ... (2)
Clearly, PQ || SR and PQ = SR [From equation (1) and (2)]
Since in quadrilateral PQRS, one pair of opposite sides are equal and parallel to each other, it is a parallelogram.
∴ PS || QR and PS = QR (Opposite sides of the parallelogram) ... (3)
In △BCD, Q and R are the mid-points of side BC and CD respectively.
∴ QR || BD and QR = 1/2 BD (Mid-point theorem) ... (4)
However, the diagonals of a rectangle are equal.
∴ AC = BD ... (5)
Thus, QR = 1/2 AC
Also, in △BAD
PS || BD and PS = 1/2 BD
Thus, QR = PS .... (6)
By using Equations (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.
