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प्रश्न
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
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उत्तर
Given: Let ABCD be a rectangle where P, Q, R, S are the midpoint of AB, BC, CD, DA.
To Prove: PQRS is a rhombus
Construction: Draw two diagonal BD and AC as shown in figure. Where BD = AC
(Since diagonal of the rectangle are equal)
Proof:
From ΔABD and ΔBCD
PS = `1/2` BD = QR and PS || BD || QR
2PS = 2QR = BD and PS || QR ...(1)
Similarly, 2PQ = 2SR = AC and PQ || SR ...(2)
From (1) and (2) we get
PQ = QR = RS = PS
Therefore, PQRS is a rhombus.
Hence, proved.
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संबंधित प्रश्न
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:
- SR || AC and SR = `1/2AC`
- PQ = SR
- PQRS is a parallelogram.
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.
In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
In the below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT
i) DP = PC (ii) PR = `1/2` AC
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral
bisect each other.
L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.
- Prove that BDFE is a parallelogram
- Find AB, if EF = 4.8 cm.
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In ΔABC, D, E and F are the midpoints of AB, BC and AC.
Show that AE and DF bisect each other.
P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.
