Question

Suppose that the width of a certain rectangle is 1 inch more than one-fourth of its length. The perimeter of the rectangle is 42 inch. How do you find the length and width of rectangle?

Answer 1

Length `16` inch, and, Width=`5` inch.

Explanation:

Let the length of the rectangle be `l` inch, so, its widh

`w=(1/4l+1)` inch.

So, the perimeter `=2l+2w=2l+2*(1/4l+1)=2l+1/2l+2=5/2l+2`.

By what is given, `5/2l+2=42 rArr 5/2l=42-2=40 rArr l=40*2/5=16`

Hence, width`=w=1/4*16+1=5`

Thus, length `16` inch, and, width=5# inch.

Answer 2

length = 16" and width = 5"

Explanation:

we can set this up with some "let x ..." statements

Let the length (L) be represented by `x`
Let the width (W) be represented by `x/4 + 1`

since the width is 1 inch longer than a quarter of it's length

the perimeter is the sum of 2 lengths and 2 widths

`2L + 2W = 42`

sub in values and solve

`2(x) + 2((x/4) + 1)`

`2x + (2x)/4 + 2 = 42`

eliminate the fraction by multiplying everything by 4

`8x + 2x + 8 = 168`

collect like terms

`10x = 160`

`:. L = 16`

sub in to find the width

`W = (16/4) + 1`

`W = 4 + 1`

`W = 5`

verify

`2L + 2W = 42`

`2(16) + 2(5) = 42`

`32 + 10 = 42`

`42 = 42`

the length is 16" and the width 5"

Answer 3

Length is 16
Width is 5

Explanation:

Breaking the question down into its component parts:

The width is 1 inch more than: `-> w=?+1`
1 fourth`" "->w=?/4+1`
of its length`" "->w=L/4+1` .................Equation(1)

The perimeter is 42 `" "-> 2w+2L=42`............Equation(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From equation(2) `w=42/2-2/2L = 21-L` ....................Equation(3)

Using equation(3) substitute for `w` in equation(1)

`color(brown)(w=L/4+1)color(blue)(" "->" "21-L=L/4+1)`

`-L-L/4=1-21`

`-5/4L=-20`

Multiply both sides by (-1)

`5/4L=20`

`color(green)(=>L=4/5xx20 = 16)`

substitute for L in equation(2)#

`2w+2L=42" "->" "2w+2(16)=42`

`color(green)(w=5)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
Using the wording of the question the following should be true:

`5->1+(1/4xx16)=5 larr "True"`