Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
EasyIf in the figure,$PQRS$ and $EFRS$ are two parallelograms,then $\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(PQRS)$. State whether the statement is True or False.
(TRUE) The statement is True.
Given that $PQRS$ and $EFRS$ are two parallelograms on the same base $SR$ and between the same parallels $PF$ and $SR$.
Therefore,$\operatorname{ar}(PQRS) = \operatorname{ar}(EFRS)$.
Now,consider the triangle $MFR$ and the parallelogram $EFRS$. They are on the same base $FR$ and between the same parallels $EM$ (which is part of $EF$) and $SR$. However,looking at the geometry,$M$ lies on $SE$. The triangle $MFR$ and parallelogram $EFRS$ share the same base $FR$ and are between the same parallels $EF$ and $SR$.
Actually,the standard theorem states that if a triangle and a parallelogram are on the same base and between the same parallels,then the area of the triangle is half the area of the parallelogram.
Here,$\triangle MFR$ and parallelogram $EFRS$ are on the same base $FR$ and between the same parallels $EF$ and $SR$.
Therefore,$\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(EFRS)$.
Since $\operatorname{ar}(EFRS) = \operatorname{ar}(PQRS)$,we have $\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(PQRS)$.
Given that $PQRS$ and $EFRS$ are two parallelograms on the same base $SR$ and between the same parallels $PF$ and $SR$.
Therefore,$\operatorname{ar}(PQRS) = \operatorname{ar}(EFRS)$.
Now,consider the triangle $MFR$ and the parallelogram $EFRS$. They are on the same base $FR$ and between the same parallels $EM$ (which is part of $EF$) and $SR$. However,looking at the geometry,$M$ lies on $SE$. The triangle $MFR$ and parallelogram $EFRS$ share the same base $FR$ and are between the same parallels $EF$ and $SR$.
Actually,the standard theorem states that if a triangle and a parallelogram are on the same base and between the same parallels,then the area of the triangle is half the area of the parallelogram.
Here,$\triangle MFR$ and parallelogram $EFRS$ are on the same base $FR$ and between the same parallels $EF$ and $SR$.
Therefore,$\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(EFRS)$.
Since $\operatorname{ar}(EFRS) = \operatorname{ar}(PQRS)$,we have $\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(PQRS)$.
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