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$ABCD$ is a parallelogram, $G$ is the point on $AB$ such that $AG = 2GB, E$ is a point of $DC$ such that $CE = 2DE$ and $F$ is the point of $BC$ such that $BF = 2FC$. Find what portion of the area of parallelogram is the area of $\triangle EFG$.
Given:
$ABCD$ is a parallelogram, $G$ is the point on $AB$ such that $AG = 2GB, E$ is a point of $DC$ such that $CE = 2DE$ and $F$ is the point of $BC$ such that $BF = 2FC$.
To do:
We have to find what portion of the area of parallelogram is the area of $AEFG$.
Solution:
Draw $EP \perp AB$ and $EQ \perp BC$
$A B=2 G B, C E=2 D E$ and $\mathrm{BF}=2 \mathrm{FC}$
This implies,
$A B-G B=2 G B$
$C D-D E=2 D E$
$\mathrm{BC}-\mathrm{FC}=2 \mathrm{FC}$
$\mathrm{AB}=3 \mathrm{BG}$ and $\mathrm{CD}=3 \mathrm{DE}$
$\mathrm{BC}=3 \mathrm{FC}$
$\mathrm{GB}=\frac{1}{3} \mathrm{AB}$, $\mathrm{DE}=\frac{1}{3} \mathrm{CD}$ and $\mathrm{FC}=\frac{1}{3} \mathrm{BC}$.........(i)
$ \operatorname{ar}(\Delta \mathrm{EFG})=\operatorname{ar}(\text { trap. BGEC })-\operatorname{ar}(\Delta \mathrm{BGF})$............(vii)
$=\frac{1}{2}\left(\frac{1}{3} \mathrm{AB}+\frac{2}{3} \mathrm{CD}\right) \times \mathrm{EP}$
$=\frac{1}{2} \mathrm{AB} \times \mathrm{EP}$
$=\frac{1}{2} \operatorname{ar}({\mathrm{parallelogram}} \mathrm{ABCD})$
$\operatorname{ar}(\Delta \mathrm{EFC})=\frac{1}{9} \operatorname{ar}({\mathrm{parallelogram}} \mathrm{ABCD})$
$\operatorname{ar}(\Delta \mathrm{BGF})=\frac{1}{2} \mathrm{BF} \times \mathrm{GR}$
$=\frac{1}{2} \times \frac{2}{3} \mathrm{BC} \times \mathrm{GR}$
$=\frac{2}{3} \times \frac{1}{2} \mathrm{BC} \times \mathrm{GR}$
$=\frac{2}{3} \times \operatorname{ar}(\Delta \mathrm{GBC})$
$=\frac{2}{3} \times \frac{1}{2} \mathrm{~GB} \times \mathrm{EP}$
$=\frac{1}{3} \times \frac{1}{3} \mathrm{AB} \times \mathrm{EP}$
$=\frac{1}{9} \mathrm{AB} \times \mathrm{EP}$
$=\frac{1}{9} \operatorname{ar}({\mathrm{parallelogram}} \mathrm{ABCD})$
$\operatorname{ar}(\Delta \mathrm{EFG})=\frac{1}{2} \operatorname{ar}({\mathrm{parallelogram}} \mathrm{ABCD})-\frac{1}{9} \operatorname{ar}(parallelogram \mathrm{ABCD})-\frac{1}{9} \operatorname{ar}({\mathrm{parallelogram}} \mathrm{ABCD})=\frac{5}{18} a r(parallelogram \mathrm{ABCD})$
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