MCQ
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are two equilateral triangles such that $D$ is the mid $-$ point of $BC$. Then, $\text{ar}(\triangle\text{BDE}):\text{ar}(\triangle\text{ABC})=?$
- A$1:2$
- ✓$1:4$
- C$\sqrt{3}:2$
- D$3:4$
✓
Answer
Correct option: B.
$1:4$
Since $D$ is the mid $-$ point of $BC,\text{BD}=\frac{\text{BC}}{2}.$
Area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text{side})^2$
$\frac{\text{ar}(\triangle\text{BDE})}{\text{ar}(\triangle\text{ABC})}=\frac{\frac{\sqrt{3}}{4}\text{BD}^2}{\frac{\sqrt{3}}{4}\text{BC}^2}$
$=\frac{\frac{\sqrt{3}}{4}\Big(\frac{\text{BC}}{2}\Big)^2}{\frac{\sqrt{3}}{4}\text{BC}^2}$
$=\frac{\frac{\text{BC}^2}{4}}{\text{BC}^2}$
$=\frac{1}{4}$
Hence, the $\text{ar}(\triangle\text{BDE}):\text{ar}(\triangle\text{ABC})$ is $1 : 4$.
Area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text{side})^2$
$\frac{\text{ar}(\triangle\text{BDE})}{\text{ar}(\triangle\text{ABC})}=\frac{\frac{\sqrt{3}}{4}\text{BD}^2}{\frac{\sqrt{3}}{4}\text{BC}^2}$
$=\frac{\frac{\sqrt{3}}{4}\Big(\frac{\text{BC}}{2}\Big)^2}{\frac{\sqrt{3}}{4}\text{BC}^2}$
$=\frac{\frac{\text{BC}^2}{4}}{\text{BC}^2}$
$=\frac{1}{4}$
Hence, the $\text{ar}(\triangle\text{BDE}):\text{ar}(\triangle\text{ABC})$ is $1 : 4$.
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