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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.
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
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Prove that : (i) AP = 2AR
(ii) BC = 4QR
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: QAP is a straight line.
The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
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(i) R is the mid-point of BC, and
(ii) PR = `(1)/(2)"DB"`.
In ΔABC, X is the mid-point of AB, and Y is the mid-point of AC. BY and CX are produced and meet the straight line through A parallel to BC at P and Q respectively. Prove AP = AQ.
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: ΔHEB ≅ ΔHFC
E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that AF = `1/3` AC.
