SOLUTION: can you please show me and explain EACH step in setting up and solving the following application problems. Give answers to the nearest thousandth if rounding is needed. c. The l

Algebra ->  Square-cubic-other-roots -> SOLUTION: can you please show me and explain EACH step in setting up and solving the following application problems. Give answers to the nearest thousandth if rounding is needed. c. The l      Log On


   

Question 39247: can you please show me and explain EACH step in setting up and solving the following application problems. Give answers to the nearest thousandth if rounding is needed.
c. The length of a rectangle is 2 cm longer than its width. If the diagonal of the rectangle is 10cm. What are the dimensions (length and width) of the rectangle?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the Pythagorean theorem.
The diagonal (D) of a rectangle is the hypotenuse of a right triangle whose legs are the rectangle's length (L) and width (W). So we can write:
D%5E2+=+L%5E2+%2B+W%5E2
But we are also given that the length (L) is 2 cm longer than its width (W), so we can write: L = W+2 Substituting this into the above equation, we have:
D%5E2+=+%28W%2B2%29%5E2+%2B+W%5E2 and we know that D = 10 cm, so...
10%5E2+=+%28W%2B2%29%5E2+%2B+W%5E2 Simplify and solve for W.
100+=+W%5E2%2B4W%2B4%2BW%5E2
100+=+2W%5E2%2B4W%2B4 Subtract 100 from both sides of the equation.
2W%5E2%2B4W-96+=+0 Divide both sides by 2 to simplify.
W%5E2%2B2W-48+=+0 Factor the quadratic equation.
%28W-6%29%28W%2B8%29+=+0 Apply the zero products principle.
W-6+=+0 and/or W%2B8+=+0
If W-6+=+0 then W+=+6
If W%2B8+=+0 then W+=+-8 Discard this solution as the width must be a positive number.
The width is 6 cm.
The length is width + 2 cm = 6 + 2 = 8 cm.