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  • 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 trapezium of area 24 cm2 (D) a rhombus of area 24 cm2

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 trapezium of area 24 cm2
(D) a rhombus of area 24 cm2

Answers (1)

Answer: [D]
Solution.
Let ABCD be a rectangle.
Given sides 8 cm and 6 cm.

Let length of rectangle = 8 cm
Breadth of rectangle = 6 cm
Let E, F, G and H are the mid-points of the sides of rectangle
AE = EB = DG = GC = 3 cm
AH = HD = BF = FC = 4 cm
Then we have to find out about the figure EFGH.
In \triangle HAE, using Pythagoras theorem
EH^{2}= AE^{2}+AH^{2}
EH^{2}=3^{2}+4^{2}=9+16=25
So, EH = 5 cm.
Similarly we can find out HG = GF = FE = 5 cm
Hence EFGH is a rhombus.
Now, area of rhombus  =  \frac{1}{2}d_{1}d_{2}
Where d_{1},d_{2}  are the diagonals of the rhombus
area of rhombus = \frac{1}{2}(EG)(HF)
Here, EG = AD = 8 cm
and HF = AB = 6 cm
\ Area of rhombus EFGH =  \frac{8\times 6}{2}=4\times 6=24cm^{2}
Therefore option (D) is correct.

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