Question 141

If $$\frac{a}{b}=\frac{25}{6}$$, then the value of $$\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$$ is

Solution

Given $$\frac{a}{b}=\frac{25}{6}$$

Squaring on both sides we get,

$$\frac{a^{2}}{b^{2}} = \frac{625}{36}$$.......(1)

We know that, $$\frac{a}{b}$$ can be written as $$\frac{a-b}{a+b}$$

$$\therefore Equation (1) can be written as,

$$\frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{625 - 36}{625 + 36}$$ = $$\frac{589}{661}$$

Hence, option A is the correct answer.

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