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Question

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $$8\ cm$$ and $$6\ cm$$ is


A
a rectangle of area 24 cm2
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B
a square of area 25 cm2
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C
a rhombus of area 24 cm2
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D
a trapezium of area 12 cm2
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Solution

The correct option is C a rhombus of area $$24\ {cm}^2$$
We know that diagonals of a rectangle are congruent. 
$$\therefore AC = BD$$
Join points $$AC$$ (diagonal)
Consider $$\Delta ABC$$
Since $$P$$ and $$Q$$ are midpoints of the two sides of the triangle,
we can say that $$PQ || AC$$ and $$PQ = \dfrac12AC $$        ..(i)
The segment joining midpoints of any $$2$$ sides of a triangle is parallel to the $$3^{rd}$$ side and is half of it
Similarly in $$\Delta ADC$$, $$RS\parallel AC$$ and $$RS = \dfrac12AC$$      ..(ii)
From (i) and (ii),  $$PQ = RS$$
Similarly, if we join $$BD$$ (diagonal), we can prove that $$PS = QR$$
Since the diagonals of rectangle are equal in length, we can say $$PQ = RS = PS = QR$$
Thus, $$ PQRS$$ is a rhombus
Area of rhombus $$= \dfrac {1}{2} \times \text{product of diagonals}$$
$$= \dfrac {1}{2} \times QS \times PR$$
$$= \dfrac {1}{2} \times 8 \times 6 = 24\ {cm}^2$$

Hence, option C.

365729_77889_ans_f67defd1f372428ebd8bd5a2c6d420c7.png

Mathematics

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