IIFT 2009 Question Paper

We know length of an arc is = angle subtended in radians * radius of the circle 

Therefore in our case Length of the arc = x*r

Also, area of sector = $$\frac{\textrm{angle subtended in radians}}{2}$$ * $$radius^2$$

Therefore in our case area of sector = $$\frac{x}{2}$$ * $$r^2$$

Also given that  area of sector = length of an arc^2

Therefore $$x^2*r^2$$ = $$\frac{x}{2}$$ * $$r^2$$

Solving we get x= 0.5

Therefore our answer is Option "A"

Let $$a, b, c$$ and $$d$$ be the sides of the cyclic quadrilateral.
Thus, $$s = \frac{(a+b+c+d)}{2}$$
$$Area = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$
Thus, after a threefold increase in the sides the sides will become $$4a, 4b, 4c$$ and $$4d$$.
Thus new $$s = s_{new} = (4a+4b+4c+4d)/2 = 4s$$
Thus, $$Area_{new} = \sqrt{(4s-4a)(4s-4b)(4s-4c)(4s-4d)}$$
= $$\sqrt{4*4*4*4*(s-a)(s-b)(s-c)(s-d)}$$  = $$16*Area$$
Thus, there is 1500% increase in the area.
Hence, option D is the correct answer.