# Ex.6.5 Q15 Triangles Solution - NCERT Maths Class 10

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## Question

In an equilateral triangle $$ABC$$, $$D$$ is a point on side $$BC$$ such that $$BD$$ $$=$$\begin{align}\frac{1}{3}BC\end{align}

Prove that $$9A{{D}^{2}}=7AB{}^{2}$$ .

Diagram

## Text Solution

Reasoning:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Steps:

$$\text{In}\,\,\Delta A B C ; A B=B C=C A \,\,\text{and}\,\,BD=\frac{1}{3}BC \\ \qquad\text{Draw} \,AE\bot BC \\\qquad BE=CE=\frac{1}{2}BC$$

[ $$\because$$ In an equilateral triangle perpendicular drawn from vertex to opposite side bisects the side]

Now In $$\Delta ADE$$

$$A{{D}^{2}}=AE{}^{2}+D{{E}^{2}}$$ (Pythagoras theorem)

\begin{align}=(\frac{\sqrt{3}}{2}BC)+{{(BE-BD)}^{2}}\end{align}

[ $$\because$$ $$AE$$ is the height of an equilateral triangle which is equal to \begin{align}\frac{\sqrt{3}}{2}\end{align} side]

\begin{align}A{D^2} &= \frac{3}{4}B{C^2} + {\left[ {\frac{{BC}}{2} - \frac{{BC}}{3}} \right]^2}\\A{D^2} &= \frac{3}{4}B{C^2} + {\left( {\frac{{BC}}{6}} \right)^2}\\ A{D^2} &= \frac{3}{4}B{C^2} + \frac{{B{C^2}}}{{36}}\\A{D^2} &= \frac{{27B{C^2} + B{C^2}}}{{36}}\\36A{D^2} &= 28B{C^2}\\9A{D^2} &= 7B{C^2}\\9A{D^2} &= 7A{B^2}\,\,\left[ {\,\,AB = BC = CA} \right]\end{align}

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