The length of a rectangle is 4 inches greater than twice the

The length of a rectangle is
4 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions of the rectangle.
bottom length of rectangle is 2w+6 and the diagonal of rectangle and hypotenuse of triangle is 2w+4w.

What is the length of the rectangle in inches? What is the width of the rectangle in inches?

Solution

Now we know that the rectangle has a length 4 more than twice its width. So in algebra this would be l = 4+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 + 4 + 2w = 6+2w.

So our equations are l = 4+2w and d = 6+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:
(4+2w)^2 + w^2 = (6+2w)^2

Now we can solve for the width.

(4+2w)^2 + w^2 = (6+2w)^2.
16 + 16w + 4w^2 + w^2 = 36 + 24w +4 w^2
w^2 -8w-20=0

w= 10,-2

take,w= 10 ,now

l=4+2w = 4+2 (10)=24 inch

d=2+l= 2+24 =26 inch.