Question

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?

Answer 1

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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