Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumWrite True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$
(A) Let the side length of the equilateral triangle $ABC$ be $x$.
Since $ABC$ is an equilateral triangle,its area is given by $ar(\triangle ABC) = \frac{\sqrt{3}}{4} x^2$.
Given that $D$ is the mid-point of $BC$,the side length of the equilateral triangle $BDE$ is $BD = \frac{BC}{2} = \frac{x}{2}$.
The area of the equilateral triangle $BDE$ is $ar(\triangle BDE) = \frac{\sqrt{3}}{4} \left(\frac{x}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{x^2}{4} = \frac{1}{4} \left(\frac{\sqrt{3}}{4} x^2\right)$.
Substituting the area of $\triangle ABC$,we get $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC)$.
Therefore,the given statement is True.
Since $ABC$ is an equilateral triangle,its area is given by $ar(\triangle ABC) = \frac{\sqrt{3}}{4} x^2$.
Given that $D$ is the mid-point of $BC$,the side length of the equilateral triangle $BDE$ is $BD = \frac{BC}{2} = \frac{x}{2}$.
The area of the equilateral triangle $BDE$ is $ar(\triangle BDE) = \frac{\sqrt{3}}{4} \left(\frac{x}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{x^2}{4} = \frac{1}{4} \left(\frac{\sqrt{3}}{4} x^2\right)$.
Substituting the area of $\triangle ABC$,we get $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC)$.
Therefore,the given statement is True.
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