  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  B
a square of area 25 cm2  C
a rhombus of area 24 cm2  D
a trapezium of area 12 cm2  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 itSimilarly 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 rhombusArea 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. Mathematics

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