The width of a rectangle is 5 less than twice its length. If the area of the rectangle is 126 cm^2, what is the length of the diagonal?

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Answer 1 · 2015-09-10T19:06:04

#sqrt(277)"cm" ~~ 16.64"cm"#

If #w# is the width of the rectangle, then we are given that:

#w(w+5) = 126#

So we would like to find a pair of factors with product #126# which differ by #5# from one another.

#126 = 2 * 3 * 3 * 7 = 14 * 9#

So the width of the rectangle is #9"cm"# and the length is #14"cm"#

Alternative method

Instead of factoring in this way, we could take the equation:

#w(w+5) = 126#

rearrange it as #w^2+5w-126 = 0#

and solve using the quadratic formula to get:

#w = (-5+-sqrt(5^2-(4xx1xx126)))/(2xx1)=(-5 +-sqrt(25+504))/2#

#=(-5+-sqrt(529))/2=(-5+-23)/2#

that is #w = -14# or #w = 9#

We are only interested in the positive width so #w = 9#, giving us the same result as the factoring.

Finding the diagnonal

Using Pythagoras theorem, the length of the diagonal in cm will be:

#sqrt(9^2+14^2) = sqrt(81+196) = sqrt(277)#

#277# is prime, so this does not simplify any further.

Using a calculator find #sqrt(277) ~~ 16.64#