Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumDiagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect at $O$ in such a way that $\operatorname{ar}(AOD) = \operatorname{ar}(BOC)$. Prove that $ABCD$ is a trapezium.
(N/A) We have a quadrilateral $ABCD$ and its diagonals $AC$ and $BD$ intersect at $O$ such that $\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$.
Now,$\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$ [Given].
Adding $\operatorname{ar}(\Delta AOB)$ to both sides,we have:
$\operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta AOB) = \operatorname{ar}(\Delta BOC) + \operatorname{ar}(\Delta AOB)$
$\Rightarrow \operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta ABC)$.
Since these two triangles $\Delta ABD$ and $\Delta ABC$ are on the same base $AB$ and have equal areas,they must lie between the same parallel lines.
Therefore,$AB \parallel DC$.
Since $ABCD$ is a quadrilateral having a pair of opposite sides parallel,$ABCD$ is a trapezium.
Now,$\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$ [Given].
Adding $\operatorname{ar}(\Delta AOB)$ to both sides,we have:
$\operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta AOB) = \operatorname{ar}(\Delta BOC) + \operatorname{ar}(\Delta AOB)$
$\Rightarrow \operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta ABC)$.
Since these two triangles $\Delta ABD$ and $\Delta ABC$ are on the same base $AB$ and have equal areas,they must lie between the same parallel lines.
Therefore,$AB \parallel DC$.
Since $ABCD$ is a quadrilateral having a pair of opposite sides parallel,$ABCD$ is a trapezium.
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Similar Questions
$XY$ is a line parallel to side $BC$ of a triangle $ABC$. If $BE || AC$ and $CF || AB$ meet $XY$ at $E$ and $F$ respectively,show that $\operatorname{ar}(ABE) = \operatorname{ar}(ACF)$.
Difficult
View SolutionThe side $AB$ of a parallelogram $ABCD$ is produced to any point $P$. $A$ line through $A$ and parallel to $CP$ meets $CB$ produced at $Q$ and then parallelogram $PBQR$ is completed. Show that $\text{ar}(ABCD) = \text{ar}(PBQR)$.
[Hint: Join $AC$ and $PQ$. Now compare $\text{ar}(ACQ)$ and $\text{ar}(APQ)$.]
[Hint: Join $AC$ and $PQ$. Now compare $\text{ar}(ACQ)$ and $\text{ar}(APQ)$.]
Difficult
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:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
Medium
View SolutionDiagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that $ar(APB) \times ar(CPD) = ar(APD) \times ar(BPC)$.
Medium
View SolutionDiagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB || DC$ intersect each other at $O$. Prove that $ar(AOD) = ar(BOC).$
Medium
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