MCQ
Medians of $\triangle\text{ABC}$ intersect at $G$. If $\text{ar}(\triangle\text{ABC})=27\text{cm}^2,$ then $\text{ar}(\triangle\text{BGC}) =$
- A$6 \mathrm{~cm}^2$
- ✓$9 \mathrm{~cm}^2$
- C$12 \mathrm{~cm}^2$
- D$18 \mathrm{~cm}^2$
✓
Answer
Correct option: B.
$9 \mathrm{~cm}^2$
$AQ, CP$ and $RB$ are medians of $\triangle\text{ABC}.$
Consider $\triangle\text{ACP}\ \&\ \triangle\text{ACQ}$
$\text{Ar}(\triangle\text{ACP})=\frac{27}{2}\text{cm}^2$ $\big[$Median divides a $\triangle$ into two equal Area$\big]$
$\text{Ar}(\triangle\text{ACQ})=\frac{27}{2}\text{cm}^2$ $\big[$Median divides a $\triangle$ into two equal Area$\big]$
$\Rightarrow\text{Ar}(\triangle\text{ACP})=\text{Ar}(\text{ACQ})$
$\text{Ar}(\triangle\text{AGC})$ is common in both the triangles.
$\Rightarrow\text{Ar}(\triangle\text{CGQ})=\text{Ar}(\triangle\text{AGP})\ ...(1)$
Similiarly $\text{Ar}\big(\triangle\text{ABR}\big)=\frac{27}{2}\text{cm}^2=\text{Ar}\big(\triangle\text{AQB}\big)$
$\text{Ar}\big(\triangle\text{ARG}\big)$ is commpn in both the triangle.
$\Rightarrow\text{Ar}\big(\triangle\text{ARG}\big)=\text{Ar}\big(\triangle\text{GQB}\big)\dots(2)$
From figure $GR, GP, GQ$ are also medians for $\triangle\text{AGC},\triangle\text{AGB }\&\triangle\text{CGB}$ respectively.
$\Rightarrow\text{Ar}\big(\triangle\text{AGC}\big)+\text{Ar}\big(\triangle\text{AGB}\big)+\text{Ar}\big(\triangle\text{CGB}\big)=27\text{cm}^2$
$\Rightarrow2\text{Ar}\big(\triangle\text{ARG}\big)+2\text{Ar}\big(\triangle\text{AGP}\big)+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$
$\Rightarrow\big(2\text{Ar}\big(\triangle\text{ARG}\big)+\text{Ar}\big(\triangle\text{AGP}\big)+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$
From equauations $(1)$ and $(2)$.
$2\big[\text{Ar}\big(\triangle\text{GQB}\big)+\text{Ar}\big(\triangle\text{AGB}\big)\big]+\text{Ar}\big(\triangle\text{CGB}\big)=27\text{cm}^2$
$\Rightarrow2\big[\text{Ar}\big(\triangle\text{BGC}\big)\big]+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$
$\Rightarrow\text{Ar}\big(\triangle\text{BGC}\big)=9\text{cm}^2$
Hence, Correct option is $(b)$.
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