The length of a rectangle is 5 cm more than 4 times its width. If the area of the rectangle is 76 cm^2, how do you find the dimensions of the rectangle to the nearest thousandth?

1 Answer
Jul 6, 2016

Width #w~=3.7785 cm#

Length #l~=20.114cm#

Explanation:

Let length #=l#, and, width #=w.#

Given that, length=5+4(width) #rArr l=5+4w...........(1)#.

Area = 76 #rArr# length x width=76 #rArr lxxw=76........(2)#

Sub.ing for#l# from #(1)# in #(2)#, we get,

#(5+4w)w=76 rArr 4w^2+5w-76=0.#

We know that the Zeroes of Quadratic Eqn. #:ax^2+bx+c=0#, are

given by, #x={-b+-sqrt(b^2-4ac)}/(2a).#

Hence, #w={-5+-sqrt(25-4*4*(-76))}/8=(-5+-sqrt(25+1216))/8#
#=(-5+-sqrt1241)/8~=(-5+-35.2278)/8#

Since #w#, width, can not be #-ve#, we can not take #w=(-5-35.2278)/8#

Therefore, width #w=(-5+35.2278)/8==30.2278/8~=3.7785 cm#

#(1)# then, gives us, length #l=5+4(3.7785)~=20.114cm#

With these dimensions, Area #=3.7785xx 20.114=76.000749 sq.cm#.

Hence, the roots satisfy the eqns.

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