QUESTION 7
Easy
O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangle PSO and PQO are equal.
SOLUTION
Consider the triangles \(PSO\) and \(PSR\).
Both have the same base \(PS\) and lie between the same parallels \(PS || QR\).
Therefore,
\(Area(△PSO) = Area(△SOR)\)
Similarly,
Consider triangles \(PQO\) and \(QOR\)
They have the same base and lie between the same parallel lines.
Therefore,
\(Area(△PQO) = Area(△QOR)\)
Also, the diagonals of the parallelogram divides it into two equal triangles. Hence,
\(Area(△PSR) = Area(△PQR)\)
Subtracting the common parts gives
\(Area(△PSO) = Area(△PQO)\).
Both have the same base \(PS\) and lie between the same parallels \(PS || QR\).
Therefore,
\(Area(△PSO) = Area(△SOR)\)
Similarly,
Consider triangles \(PQO\) and \(QOR\)
They have the same base and lie between the same parallel lines.
Therefore,
\(Area(△PQO) = Area(△QOR)\)
Also, the diagonals of the parallelogram divides it into two equal triangles. Hence,
\(Area(△PSR) = Area(△PQR)\)
Subtracting the common parts gives
\(Area(△PSO) = Area(△PQO)\).
Concept Note
Triangles on the same base and between the same parallel lines have equal areas.
In a parallelogram,
1. Opposite sides are equal.
2. Opposite sides are parallel.