Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn a triangle $ABC$,$E$ is the mid-point of median $AD$. Show that $\operatorname{ar}(BED) = 1/4 \operatorname{ar}(ABC)$.
(N/A) We have a $\Delta ABC$ and its median $AD$.
Since a median divides the triangle into two triangles equal in area,
$\therefore \operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(\Delta ABC) \quad \dots(1)$
Now,in $\Delta ABD$,$BE$ is a median because $E$ is the mid-point of $AD$.
$\therefore \operatorname{ar}(\Delta BED) = \frac{1}{2} \operatorname{ar}(\Delta ABD) \quad \dots(2)$
From $(1)$ and $(2)$,we have:
$\operatorname{ar}(\Delta BED) = \frac{1}{2} \left[ \frac{1}{2} \operatorname{ar}(\Delta ABC) \right]$
$\Rightarrow \operatorname{ar}(\Delta BED) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$
Since a median divides the triangle into two triangles equal in area,
$\therefore \operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(\Delta ABC) \quad \dots(1)$
Now,in $\Delta ABD$,$BE$ is a median because $E$ is the mid-point of $AD$.
$\therefore \operatorname{ar}(\Delta BED) = \frac{1}{2} \operatorname{ar}(\Delta ABD) \quad \dots(2)$
From $(1)$ and $(2)$,we have:
$\operatorname{ar}(\Delta BED) = \frac{1}{2} \left[ \frac{1}{2} \operatorname{ar}(\Delta ABC) \right]$
$\Rightarrow \operatorname{ar}(\Delta BED) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$
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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:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
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$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
Medium
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