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Areas of Parallelograms and Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreas of Parallelograms and Triangles1 Mark
Question
Write True or False and justify your answer: $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\text{ar}(\triangle\text{BDE})=\frac{1}{4}\text{ar}(\triangle\text{ABC}).$
✓

Answer

Given, $\triangle\text{ABC}$ and $\triangle\text{BDE}$ are two equilateral triangle.
$\therefore$ Area of an equilateral
$\triangle\text{ABC}=\frac{\sqrt{3}}{4}\times(\text{sides})^2=\frac{\sqrt{3}}{4}(\text{BC})^2$
$[\because$ in equilateral $\triangle\text{ABC}, Ab = BC = AC]$
Also, given $D$ is the mid point of $BC.$
$\therefore\text{BD}=\text{DC}=\frac{1}{2}\text{BC}\ ...(\text{ii})$
Now, area of an equilateral $\triangle\text{BDE}=\frac{\sqrt{3}}{4}\times(\text{sides})^2$
$=\frac{\sqrt{3}}{4}\times(\text{BD})^2 [ \therefore$ in equilateral $\triangle\text{BDE} , BD = DE = BE]$
$=\frac{\sqrt{3}}{4}\times\Big(\frac{1}{2}\text{BC}\Big)^2 [$from eq. $(ii) =\frac{\sqrt{3}}{4}\times\frac{1}{4}\text{BC}^2=\frac{1}{4}\Big(\frac{\sqrt{3}}{4}\text{BC}^2\Big)$
$\text{Area of}\ \triangle\text{(BDE} )=\frac{1}{4}\text{Area of}\ \triangle\text{ABC}.$

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