Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn the figure,$BD \parallel CA$,$E$ is the mid-point of $CA$,and $BD = \frac{1}{2} CA$. Prove that $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.
(N/A) Given: $BD \parallel CA$,$E$ is the mid-point of $CA$,so $CE = \frac{1}{2} CA$.
Since $BD = \frac{1}{2} CA$,we have $BD = CE$.
Also,$BD \parallel CE$ (as $BD \parallel CA$).
Since one pair of opposite sides is equal and parallel,$BCED$ is a parallelogram.
Now,$\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$ because they lie on the same base $BC$ and between the same parallels $BC$ and $DE$.
In $\triangle ABC$,$BE$ is a median because $E$ is the mid-point of $CA$.
$A$ median of a triangle divides it into two triangles of equal area.
Therefore,$\operatorname{ar}(EBC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(EBC)$.
Substituting $\operatorname{ar}(EBC) = \operatorname{ar}(DBC)$,we get $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.
Hence proved.
Since $BD = \frac{1}{2} CA$,we have $BD = CE$.
Also,$BD \parallel CE$ (as $BD \parallel CA$).
Since one pair of opposite sides is equal and parallel,$BCED$ is a parallelogram.
Now,$\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$ because they lie on the same base $BC$ and between the same parallels $BC$ and $DE$.
In $\triangle ABC$,$BE$ is a median because $E$ is the mid-point of $CA$.
$A$ median of a triangle divides it into two triangles of equal area.
Therefore,$\operatorname{ar}(EBC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(EBC)$.
Substituting $\operatorname{ar}(EBC) = \operatorname{ar}(DBC)$,we get $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.
Hence proved.
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Similar Questions
In $\Delta ABC$,$AD$ is a median and $AM$ is an altitude. The side $BA$ of $\Delta ABC$ is produced to any point $E$,so that $AB = AE$. If $BC = 16\, cm$ and $AM = 8\, cm$,then find the area of $\Delta EBD$ in $cm^2$.
Medium
View Solution$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ $(Fig.)$. $AE$ intersects $CD$ at $F.$ If $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,find the area of the parallelogram $ABCD$ (in $\text{cm}^2$).
Difficult
View Solution$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$
Difficult
View SolutionIn quadrilateral $ABCD$,$AM$ and $CN$ are altitudes on diagonal $BD$ drawn from $A$ and $C$ respectively. Prove that,$\operatorname{ar}(ABCD) = \frac{1}{2} \times BD \times (AM + CN)$.
Medium
View SolutionWrite True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC).$
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC).$
Medium
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