If $$(x+y):(x-y)=3:2$$, then $$(x^2+y^2):(x^2-y^2)$$ is in the ratio of:
Given,
$$\frac{(x+y)}{(x-y)}=\frac{3}{2}$$,
 Using componendo and dividendo rule,
$$\frac{(x+y)+(x-y)}{(x+y)-(x-y)}=\frac{3+2}{3-2}$$
$$\frac{x}{y}=\frac{5}{1}$$
Putting the values in $$\frac{(x^2+y^2)}{(x^2-y^2)}$$ = $$\frac{(5^2+1^2)}{(5^2-1^2)}$$=$$\frac{26}{24}=13:12$$
Option C is correct.
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