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प्रश्न
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.
विकल्प
a rhombus
a rectangle
a square
any parallelogram
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उत्तर
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is a rectangle.
Explanation:
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
Join AC, RP and SQ
In ∆ABC,
P is midpoint of AB and Q is midpoint of BC
∴ By midpoint theorem,
PQ || AC and PQ = `1/2` AC ...(1)
Similarly,
In ∆DAC,
S is midpoint of AD and R is midpoint of CD
∴ By midpoint theorem,
SR || AC and SR = `1/2` AC ...(2)
From (1) and (2),
PQ || SR and PQ = SR
⇒ PQRS is a parallelogram
ABQS is a parallelogram
⇒ AB = SQ
PBCR is a parallelogram
⇒ BC = PR
⇒ AB = PR ...[∵ BC = AB, sides of rhombus]
⇒ SQ = PR
∴ Diagonals of the parallelogram are equal.
Hence, it is a rectangle.
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संबंधित 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.
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.
In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.
[Hint: DN || QM]
In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF.
Show that: EF = AC.
In ∆ABC, E is the mid-point of the median AD, and BE produced meets side AC at point Q.
Show that BE: EQ = 3: 1.
In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
In ΔABC, D, E and F are the midpoints of AB, BC and AC.
Show that AE and DF bisect each other.
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.
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).
[Hint: Join BE and produce it to meet CD produced at G.]
