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1. Hi, could someone explain to me why my approach to the following question doesn't work? Three squares are arranged as shown so that their baseslie on a straight line. Also, the corners P, Q and R lieon a straight line. The middle square has sides that are8 cm longer than the sides of the smallest square. Thelargest square has sides of length 50 cm. There are twopossible values for the length (in cm) of the sides of thesmallest square.

https://www.ukmt.org.uk/pdfs/SMC2015_sol_extended.pdf

question 19 in the solutions, there is a diagram.

I started by calculating the gradient of the first point to the second point...

which is 8/x

I then calculated the gradient from the first point to the last point...

which is (50-x)/(x+8)

I then equated the two gradients...

8/x = (50-x)/(x+8)

8 = x(50-x)/(x+8)

8x + 64 = 50x - x^2

x^2 - 42x + 64 = 0

However this quadratic is not soluble without a calculated (at least by me)

Thanks for any help
2. (Original post by staircase)

I then calculated the gradient from the first point to the last point...

which is (50-x)/(x+8)
It doesn't work because of that.

The vertical change is indeed 50-x between the largest and smallest squares, however the horizontal change is x+(x+8) because (x+8) on its own only accounts for the length of the middle square whereas you need the smaller square as well, which is the x.

Updated: October 16, 2016
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