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?

3 Answers
Aug 7, 2016

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.

Aug 7, 2016

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"

Aug 7, 2016

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"#