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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.
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
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.
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
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In ΔABC, D is the mid-point of AB and E is the mid-point of BC.
Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 72°
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