There are two squares S 1 and S 2 with areas 8 and 9 units, respectively. S 1 is inscribed within S 2 , with one corner of S 1 on each side of S 2 . The corners of the smaller square divides the sides of
the bigger square into two segments, one of length ‘a’ and the other of length ‘b’, where, b > a. A possible value of ‘b/a’, is:

Area of $$S_1 = 8$$ sq. units

=> Side of $$S_1 = PS = \sqrt{8} = 2 \sqrt{2}$$ units

Similarly, Side of $$S_2 = CD = \sqrt{9} = 3$$ units

=> $$a + b = 3$$

In $$\triangle$$ PDS

=> $$b^2 + a^2 = 8$$

=> $$b^2 + (3 - b)^2 = 8$$

=> $$b^2 + 9 + b^2 - 6b = 8$$

=> $$2b^2 - 6b + 1 = 0$$

=> $$b = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm \sqrt{28}}{4}$$

=> $$b = \frac{3 + \sqrt{7}}{2}$$     $$(\because b > a)$$

=> $$a = 3 - \frac{3 + \sqrt{7}}{2} = \frac{3 - \sqrt{7}}{2}$$

$$\therefore \frac{b}{a} = \frac{\frac{3 + \sqrt{7}}{2}}{\frac{3 - \sqrt{7}}{2}}$$

= $$\frac{3 + \sqrt{7}}{3 - \sqrt{7}} \approx 15.9$$