Advertisements
Advertisements
प्रश्न
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Advertisements
उत्तर

Let us join AC and BD.
In ΔABC,
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 ΔADC,
SR || AC and SR = `1/2 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 || QR and PS = QR ...(Opposite sides of the parallelogram) ...(3)
In ΔBCD, 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)
By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.
APPEARS IN
संबंधित प्रश्न
In a ∆ABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ∆DEF.
Let Abc Be an Isosceles Triangle in Which Ab = Ac. If D, E, F Be the Mid-points of the Sides Bc, Ca and a B Respectively, Show that the Segment Ad and Ef Bisect Each Other at Right Angles.
Fill in the blank to make the following statement correct:
The figure formed by joining the mid-points of consecutive sides of a quadrilateral is
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)

A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.

Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
In triangle ABC, the medians BP and CQ are produced up to points M and N respectively such that BP = PM and CQ = QN. Prove that:
- M, A, and N are collinear.
- A is the mid-point of MN.
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find FE, if BC = 14 cm
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: F is the mid-point of BC.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.
