The figure obtained by joining the mid-points | vedclass

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(D) Let $ABCD$ be a rectangle with sides $AB = 8 \, cm$ and $BC = 6 \, cm$.Let $E, F, G,$ and $H$ be the mid-points of sides $AB, BC, CD,$ and $DA$ re

Vedclass · Accepted Answer

a rhombus of area $24 \, cm^{2}$

Vedclass · Answer

(D) Let $ABCD$ be a rectangle with sides $AB = 8 \, cm$ and $BC = 6 \, cm$.Let $E, F, G,$ and $H$ be the mid-points of sides $AB, BC, CD,$ and $DA$ respectively.By joining these mid-points,we obtain a quadrilateral $EFGH$.According to the mid-point theorem,the quadrilateral formed by joining the mid-points of the sides of a rectangle is a rhombus.The diagonals of this rhombus $EFGH$ are equal to the sides of the rectangle,i.e.,$EG = BC = 6 \, cm$ and $HF = AB = 8 \, cm$.The area of the rhombus is given by $\frac{1}{2} 	imes 	ext{product of diagonals} = \frac{1}{2} 	imes EG 	imes HF$.Area $= \frac{1}{2} 	imes 6 \, cm 	imes 8 \, cm = 24 \, cm^{2}$.Thus,the figure obtained is a rhombus of area $24 \, cm^{2}$.

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

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