MCQ
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are two equilateral triangles such that $D$ is the mid $-$ point of $BC$. The ratio of the areas of triangle $\text{ABC}$ and $\text{BDE}$ is :
- A$2 : 1$
- B$1 : 2$
- ✓$4 : 1$
- D$1 : 4$
✓
Answer
Correct option: C.
$4 : 1$
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are equilateral triangles and $D$ is the mid $-$ point of $PC.$
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are both equilateral triangles
$\therefore$ They are similar also
$\therefore\frac{\text{area of }\triangle\text{ABC}}{\text{area of }\triangle\text{BDE}}=\frac{\text{BC}^2}{\text{BD}^2}$
$=\frac{\text{BC}^2}{\big(\frac{1}{2}\text{BC}^2\big)} \ [D $ is mid point of $BC]$
$=\frac{\text{BC}^2}{\frac{1}{4}\text{BC}^2}=\frac{\text{4BC}^2}{\text{BC}^2}=\frac{4}{1}$
$\therefore$ Ratio is $4 : 1$
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are both equilateral triangles
$\therefore$ They are similar also
$\therefore\frac{\text{area of }\triangle\text{ABC}}{\text{area of }\triangle\text{BDE}}=\frac{\text{BC}^2}{\text{BD}^2}$
$=\frac{\text{BC}^2}{\big(\frac{1}{2}\text{BC}^2\big)} \ [D $ is mid point of $BC]$
$=\frac{\text{BC}^2}{\frac{1}{4}\text{BC}^2}=\frac{\text{4BC}^2}{\text{BC}^2}=\frac{4}{1}$
$\therefore$ Ratio is $4 : 1$
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