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?

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Answer 1 · 2016-07-06T17:07:11

Width $w ≅ 3.7785 c m$ $w~=3.7785 cm$

Length $l ≅ 20.114 c m$ $l~=20.114cm$

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

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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