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In an equilateral triangle with side a, prove that area = \frac{\sqrt{3}}{4} a^{2}
CBSE | Class 10 | Exercise 4E | Guides | Maths | RS Aggarwal | Triangles
In an equilateral triangle with side a, prove that area = \frac{\sqrt{3}}{4} a^{2}

 

 

 

 

Answer:

We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.

ABC is an equilateral triangle having AB =BC = CA = a.

AD is the altitude drawn from the vertex A to the side BC.

It will bisect the side BC.

D C=\frac{1}{2} a

In right triangle ADC

By using Pythagoras theorem,

\begin{aligned} A C^{2} &=C D^{2}+D A^{2} \\ \Rightarrow & a^{2}-\left(\frac{1}{2} a\right)^{2}+D A^{2} \\ \Rightarrow D A^{2} &=a^{2}-\frac{1}{4} a^{2} \\ \Rightarrow D A^{2} &=\frac{3}{4} a^{2} \\ \Rightarrow D A=\frac{\sqrt{3}}{4} a^{2} \\ \Rightarrow \text { Area }(\triangle A B C)=\frac{1}{2} \times B C \times A D \\ & \Rightarrow \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a \\ & \Rightarrow \frac{\sqrt{3}}{4} a^{2} \end{aligned}

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