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AreaAreas of Parallelograms and Triangles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSAreaAreas of Parallelograms and Triangles1 Mark
Question
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are two equilateral triangles such that D is the mid-point of BC. Then, $\text{ar}(\triangle\text{BDE}):\text{ar}(\triangle\text{ABC})=?$
  1. $1:2$
  2. $1:4$
  3. $\sqrt{3}:2$
  4. $3:4$
✓

Answer

  1. $1:4$
Solution:
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.

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