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 ABC and BDE is:
- A2 : 1
- B1 : 2
- ✓4 : 1
- D1 : 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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