The length of a rectangle is 59 inches greater than twice the width. if the diagonal is 2 inches more than the length, find the dimensions of the rectangle

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 20 more than twice its width. So in algebra this would be l = 20+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 + 20 + 2w = 22+2w.

So our equations are l = 20+2w and d = 22+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.

(20+2w)^2 + w^2 = (22+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  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.

Dimensions of rectangle are 20 inches and 99 inches.

Explanation:

Let the width be w,

then as length, which is 59 inches greater than twice the width, is 2w+59 inches

and square of diagonal is w2+(2w+59)2=w2+4w2+236w+3481

= 5w2+236w+3481

But as diagonal is 2 more than length, it is 2w+61 and hence

5w2+236w+3481=(2w+61)2=4w2+244w+3721

Hence w2−8w−240−0 or (w−20)(w+12)=0

Hence, width is 20 or −12, but as it cannot be negative, it is 20 and length is 20×2+59=99.

Hence dimensions of rectangle are 20 inches and 99 inches.

Check 202+992=400+9801=10201=1012 and diagonal is 101 and 2 more than length.

I hope this help