1. 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:
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By: anil on 05 May 2019 02.37 am
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 $$ riangle$$ PDS => $$b^2 + a^2 = 8$$ => $$b^2 + (3 - b)^2 = 8$$ => $$b^2 + 9 + b^2 - 6b = 8$$ => $$2b^2 - 6b + 1 = 0$$ => $$b = frac{6 pm sqrt{36 - 8}}{4} = frac{6 pm sqrt{28}}{4}$$ => $$b = frac{3 + sqrt{7}}{2}$$ $$(ecause b > a)$$ => $$a = 3 - frac{3 + sqrt{7}}{2} = frac{3 - sqrt{7}}{2}$$ $$ herefore frac{b}{a} = frac{frac{3 + sqrt{7}}{2}}{frac{3 - sqrt{7}}{2}}$$ = $$frac{3 + sqrt{7}}{3 - sqrt{7}} approx 15.9$$
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