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(B) Let the angle between the given pair of lines be $	heta$. The area of a sector with angle $\alpha$ is given by $\frac{1}{2}r^2\alpha$.Given that

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$\frac{\sqrt{3}}{2}$

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(B) Let the angle between the given pair of lines be $	heta$. The area of a sector with angle $\alpha$ is given by $\frac{1}{2}r^2\alpha$.Given that the area of the bigger sector is $5$ times the area of the smaller sector:$\frac{1}{2}r^2(\pi-	heta) = 5 	imes \frac{1}{2}r^2	heta$$\pi-	heta = 5	heta$$6	heta = \pi \implies 	heta = \frac{\pi}{6}$.The angle $	heta$ between the lines represented by $ax^2+2hxy+by^2=0$ satisfies $\cos 	heta = \frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$.Substituting $	heta = \frac{\pi}{6}$:$\cos \frac{\pi}{6} = \frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$$\frac{\sqrt{3}}{2} = \frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$.

List-$I$	List-$II$
$(A)$ $5x^2 + 2\sqrt{7}xy - y^2 = 0$	$(I)$ $\frac{\sqrt{3}}{2}$
$(B)$ $x^2 + \sqrt{11}xy + 2y^2 = 0$	$(II)$ $\frac{1}{2\sqrt{3}}$
$(C)$ $x^2 + 2\sqrt{2}xy + y^2 = 0$	$(III)$ $\frac{1}{2}$
$(D)$ $3x^2 + 4\sqrt{2}xy + y^2 = 0$	$(IV)$ $\frac{2}{3}$
	$(V)$ $\frac{1}{\sqrt{2}}$

The combined equation of the two diameters of a circle which divide the circle into $4$ sectors is $ax^2+2hxy+by^2=0$. If the area of the bigger sector is $5$ times the area of the smaller sector,then $\frac{|a+b|}{\sqrt{(a-b)^2+4h^2}} = $

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