Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$PQRS$ and $ABRS$ are parallelograms and $X$ is any point on side $BR$. Show that:
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$
(N/A) We have two parallelograms $PQRS$ and $ABRS$ and a point $X$ on $BR$.
$(i)$ To prove that $ar(PQRS) = ar(ABRS)$:
Since parallelogram $PQRS$ and parallelogram $ABRS$ are on the same base $RS$ and between the same parallels $RS$ and $PB$,their areas are equal.
Therefore,$ar(PQRS) = ar(ABRS)$.
$(ii)$ To prove that $ar(AXS) = 1/2 \, ar(PQRS)$:
Since $\Delta AXS$ and parallelogram $ABRS$ are on the same base $AS$ and between the same parallels $AS$ and $BR$,the area of the triangle is half the area of the parallelogram.
Therefore,$ar(AXS) = 1/2 \, ar(ABRS)$.
Since $ar(PQRS) = ar(ABRS)$ (from part $i$),
Substituting this into the equation,we get:
$ar(AXS) = 1/2 \, ar(PQRS)$.
$(i)$ To prove that $ar(PQRS) = ar(ABRS)$:
Since parallelogram $PQRS$ and parallelogram $ABRS$ are on the same base $RS$ and between the same parallels $RS$ and $PB$,their areas are equal.
Therefore,$ar(PQRS) = ar(ABRS)$.
$(ii)$ To prove that $ar(AXS) = 1/2 \, ar(PQRS)$:
Since $\Delta AXS$ and parallelogram $ABRS$ are on the same base $AS$ and between the same parallels $AS$ and $BR$,the area of the triangle is half the area of the parallelogram.
Therefore,$ar(AXS) = 1/2 \, ar(ABRS)$.
Since $ar(PQRS) = ar(ABRS)$ (from part $i$),
Substituting this into the equation,we get:
$ar(AXS) = 1/2 \, ar(PQRS)$.
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Similar Questions
In the figure,diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at $O$ such that $OB = OD$. If $AB = CD$,then show that:
$(i)$ $ar(DOC) = ar(AOB)$
$(ii)$ $ar(DCB) = ar(ACB)$
$(iii)$ $DA \parallel CB$ or $ABCD$ is a parallelogram.
[Hint: From $D$ and $B$,draw perpendiculars to $AC$.]
$(i)$ $ar(DOC) = ar(AOB)$
$(ii)$ $ar(DCB) = ar(ACB)$
$(iii)$ $DA \parallel CB$ or $ABCD$ is a parallelogram.
[Hint: From $D$ and $B$,draw perpendiculars to $AC$.]
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View SolutionIn the figure,$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F$,show that:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
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