Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$ABC$ and $ABD$ are two triangles on the same base $AB$. If line segment $CD$ is bisected by $AB$ at $O$,show that $\operatorname{ar}(ABC) = \operatorname{ar}(ABD)$.
(N/A) We have $\Delta ABC$ and $\Delta ABD$ on the same base $AB$.
Since $CD$ is bisected at $O$,we have $CO = DO$.
In $\Delta ACD$,$AO$ is a median because it divides the side $CD$ into two equal parts $(CO = DO)$.
Therefore,$\operatorname{ar}(\Delta AOC) = \operatorname{ar}(\Delta AOD)$ (Since a median of a triangle divides it into two triangles of equal areas) $(1)$.
Similarly,in $\Delta BCD$,$BO$ is a median.
Therefore,$\operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta BOD)$ $(2)$.
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta AOC) + \operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta BOD)$
$\Rightarrow \operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta ABD)$.
Since $CD$ is bisected at $O$,we have $CO = DO$.
In $\Delta ACD$,$AO$ is a median because it divides the side $CD$ into two equal parts $(CO = DO)$.
Therefore,$\operatorname{ar}(\Delta AOC) = \operatorname{ar}(\Delta AOD)$ (Since a median of a triangle divides it into two triangles of equal areas) $(1)$.
Similarly,in $\Delta BCD$,$BO$ is a median.
Therefore,$\operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta BOD)$ $(2)$.
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta AOC) + \operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta BOD)$
$\Rightarrow \operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta ABD)$.
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Similar Questions
In the figure,$ABCDE$ is a pentagon. $A$ line through $B$ parallel to $AC$ meets $DC$ produced at $F$. Show that:
$(i)$ $ar(ACB) = ar(ACF)$
$(ii)$ $ar(AEDF) = ar(ABCDE)$
$(i)$ $ar(ACB) = ar(ACF)$
$(ii)$ $ar(AEDF) = ar(ABCDE)$
Medium
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,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(BCED) = \operatorname{ar}(ABMN) + \operatorname{ar}(ACFG)$
Medium
View SolutionIn the figure,$P$ is a point in the interior of a parallelogram $ABCD$. Show that:
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]
Difficult
View Solution$P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP$. Show that $\operatorname{ar}(RQC) = \frac{3}{8} \operatorname{ar}(ABC)$.
Medium
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