Division of length segments of a square by an inscribed square.

Question

There are two squares S1 and S2 with areas 8 and 9 units respectively. S1 is inscribed within S2, with one corner of S1 on each side of S2. 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?

 

Options:

1) >=5 and <8

2) >=8 and <11

3) >=11 and <14

4) >=14 and <17

5) >17

Answer 1

It appears that it is possible to solve for  b directly.  The side of the larger square is 3, the side of the smaller square is sqrt(8).   Study of a diagram and application of the Pythagorean theorem leads to the equation 

 

     b² + ( 3- b)² = 8   this quadratic equation can be rearranged into the standard form and solved using the quadratic formula.   The result is   b =  3/2 ± sqrt(7) /2  .   Since b > a, the plus sign must be taken.   So b ~ 2.822.

 

So the answer is  4)