# The Center of Math Blog

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## Monday, July 27, 2015

### Problem of the Week

Here's a new Problem of the Week from the Center of Math, which is a bit harder than some other ladders-resting-on-walls problems you might know.

Figured out the answer? Find out if you have it right  with the solution below...

Did you find the answer in a different way? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week.

#### 1 comment:

1. I solved it by breaking up the problem into the 2 small triangles and calling the angle in the bottom right of both triangles A and the breaking up the ladder into 35-x and x. So sin(A)=12/x and cos(A)=12/(35-x). Then using sin^2(A) + cos^2(A) =1 getting the equation 176400-10080x-(937x^2)+(70x^3)-(x^4) which factorises into -(x-15)(x-20)(x^2-35x-588). I disregard the imaginary part and the -1 as i have no need for it in my answer. Then using Pythagoras on the smallest the triangle in the diagram, i get (y^2)+(12^2)=(x^2), y being the term that i want. I then sub in the factors, x=15 and x=20 and you get the values of y as 9 and 16 respectively.