Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
DifficultIf a triangle and a parallelogram are on the same base and between the same parallels,then prove that the area of the triangle is equal to half the area of the parallelogram.
(N/A) Let $\Delta ABP$ and parallelogram $ABCD$ be on the same base $AB$ and between the same parallels $AB$ and $PC$.
You wish to prove that $\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABCD)$.
Draw $BQ \parallel AP$ to obtain another parallelogram $ABQP$. Now,parallelograms $ABQP$ and $ABCD$ are on the same base $AB$ and between the same parallels $AB$ and $PC$.
Therefore,$\text{ar}(ABQP) = \text{ar}(ABCD)$ $(1)$.
But $\Delta PAB \cong \Delta BQP$ (Diagonal $PB$ divides parallelogram $ABQP$ into two congruent triangles).
So,$\text{ar}(PAB) = \text{ar}(BQP)$ $(2)$.
Therefore,$\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABQP)$ [From $(2)$] $(3)$.
This gives $\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABCD)$ [From $(1)$ and $(3)$].
You wish to prove that $\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABCD)$.
Draw $BQ \parallel AP$ to obtain another parallelogram $ABQP$. Now,parallelograms $ABQP$ and $ABCD$ are on the same base $AB$ and between the same parallels $AB$ and $PC$.
Therefore,$\text{ar}(ABQP) = \text{ar}(ABCD)$ $(1)$.
But $\Delta PAB \cong \Delta BQP$ (Diagonal $PB$ divides parallelogram $ABQP$ into two congruent triangles).
So,$\text{ar}(PAB) = \text{ar}(BQP)$ $(2)$.
Therefore,$\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABQP)$ [From $(2)$] $(3)$.
This gives $\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABCD)$ [From $(1)$ and $(3)$].
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