Ex.6.5 Q15 Triangles Solution - NCERT Maths Class 10


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

 

 Video Solution
Triangles
Ex 6.5 | Question 15

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:

\(\begin{align} \text{In}\,\,\Delta A B C ; & \,\, A B=B C=C A \\ & \text{and}\,\,BD=\frac{1}{3}BC \\ 
 & \text{Draw} \,AE\bot BC \\ &  BE=CE=\frac{1}{2}BC \end{align} \)

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

Now In \(\Delta ADE\)

\( \begin{align} A{{D}^{2}} & =AE{}^{2}+D{{E}^{2}} \\ & [ \text { Pythagoras }  \text{ theorem } ] \end{align} \)

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

[ \(\because\) \(AE\) 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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