SOLUTION: The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 101 cm2, what is the length of the diagonal?

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Question 1125597: The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 101 cm2, what is the length of the diagonal?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let the length be L, and the width W
if the width of a rectangle is 9 less than twice its length,we have
W=2L-9....eq.1

if the area of the rectangle is A=101cm%5E2, we have
L%2AW=101cm%5E2....eq.2
substitute W from eq.1

L%282L-9%29=101cm%5E2...solve for L

2L%5E2-9L=101cm%5E2
2L%5E2-9L-101cm%5E2=0

use quadratic formula

L+=+%28-%28-9%29+%2B-+sqrt%28+%28-9%29%5E2-4%2A2%2A%28-101%29+%29%29%2F%282%2A2%29+
L+=+%289+%2B-+sqrt%2881%2B808+%29%29%2F4+

L+=+%289+%2B-+sqrt%28889+%29%29%2F4+
since length, we need only positive solution
L+=+%289+%2B+sqrt%28889+%29%29%2F4+
L+=+%289+%2B+29.8161%29%2F4+
L+=+38.8161%2F4+
L+=+9.704025+

now find W
W=2%2A9.704025-9....eq.1
W=19.40805-9
W=10.40805

check the area:
L%2AW=101cm%5E2
10.40805%2A9.704025=101cm%5E2
100.99997740125=101cm%5E2...round it
101=101cm%5E2