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Question
ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.
Prove that area of triangle APQ = `1/8` of the area of parallelogram ABCD.
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Solution
We have to join PD and BD.
BD is the diagonal of the parallelogram ABCD. Therefore it divides the parallelogram into two equal parts.
∴ Area( ΔABD )= Area ( ΔDBC )
=`1/2` Area ( parallelogram ABCD) ...(i)
DP is the median of ΔABD. Therefore it will divide ΔABD into two triangles of equal areas.
∴ Area( ΔAPD )= Area ( ΔDPB )
= `1/2` Area ( ΔABD )
= `1/2 xx 1/2` Area (parallelogram ABCD) ...[from equation (i)]
= `1/4` Area (parallelogram ABCD) ...(ii)
In ΔAPD, Q is the mid-point of AD. Therefore PQ is the median.
∴ Area(ΔAPQ)= Area (ΔDPQ)
= `1/2` Area (ΔAPD)
= `1/2 xx 1/4` Area (parallelogram ABCD)...[from equation (ii)]
Area (ΔAPQ)= `1/8` Area (parallelogram ABCD),
hence proved
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