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  • If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.).

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig.).

Answers (1)

Solution.
Given that if the mid-points of the sides of a quadrilateral are joined in order, a parallelogram is formed. We have to find the area of this parallelogram.
To prove: ar (parallelogram PFRS) =\frac{1}{2} ar(quadrilateral ABCD)

Let ABCD is a quadrilateral and P, F, R and S are the mid-points of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.
Construction:- Join BD and BR.
We know that median of a triangle divides it into two triangles of equal area.
So, BR divides \triangle BDA into two triangles of equal area.
\therefore ar(\triangle BRA) =\frac{1}{2} ar(\triangle BDA)                    …(i)
Similarly, median RS divides \triangle BRA into two triangles of equal area.
\therefore ar(\triangle ASR)=\frac{1}{2}ar(\triangle BRA)                     …(ii)
From eq. (i) and (ii) 
\therefore ar(\triangle ASR)=\frac{1}{4}ar(\triangle BDA)                    …(iii)
Similarly,
ar(\triangle CFP)=\frac{1}{4}ar(\triangle BCD)                       ….(iv)
On adding equations (iii) and (iv), we get
ar(\triangle ASR)+ar(\triangle CFP)=\frac{1}{4}ar(\triangle BDA+\triangle BCD)
\Rightarrow ar(\triangle ASR)+ar(\triangle CFP)=\frac{1}{4}ar(Quadrilateral BCDA) …(v)
Similarly,
ar(\triangle DRF)+ar(\triangle BSP)=\frac{1}{4}ar(BCDA)….(vi)
On adding (v) and (vi), we get

ar(\triangle ASR)+ar(\triangle CFP)+ar (\triangle BSP)+ar(\triangle BSP)    
=\frac{1}{2}ar (quadrilateral BCDA)                                                          …(vii)
But
ar(\triangle ASR) + ar(\triangle CFP) + ar(\triangle DRF) + ar(\triangle BSP) = ar(\parallel gm PFRS)
=\frac{1}{2}ar       (quadrilateral BCDA)                                                               …(viii)
from subtracting eq. (vii) from eq. (viii) we get
ar(parallelogram PFRS)  =\frac{1}{2}ar (quadrilateral BCDA)

Hence proved

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