word problem for mathematics

label Mathematics
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May 20th, 2015

To start we need to assign variables for what were trying to find. Lets say the rectangle has width w. Now we know that the rectangle has a length 11 more than twice its width. So in algebra this would be l =

11+2w. Now we know that the diagonal is 2 more than the length. In algebra this would be d = 2+l. Substituting our length equation into the diagonal expression we get d = 2 + 11 + 2w = 14+2w.

So our equations are l = 11+2w and d = 14+2w. We also know that the angles of rectangle are right angles. So the length, width, and diagonal form a right triangle and using the Pythagorean theorem we find that:

(20+2w)^2 + w^2 = (22+2w)^2.


Now we can solve for the width.

(11+2w)^2 + w^2 = (11+2w)^2.
400 + 80w + 4w^2 + w^2 = 484 + 44w + w^2

I will leave the rest to you but just a hint, you will have to use the quadratic formula. You will eventually get it down to w+=+%28-9+%2B-+sqrt%28+165%29%29%2F%282%29+ which if you plug into your calculator should come out to 1.9226 or -10.9226. 

Since we know that a measure of a side can't be negative we know that the width has to be 1.9226. So now we go back and find the length and the diagonal. I will leave this for you to do. You should get L = 23.8452 and D = 25.8452.

May 20th, 2015

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