# 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?

Aug 7, 2016

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

#### Explanation:

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

$w = \left(\frac{1}{4} l + 1\right)$ inch.

So, the perimeter $= 2 l + 2 w = 2 l + 2 \cdot \left(\frac{1}{4} l + 1\right) = 2 l + \frac{1}{2} l + 2 = \frac{5}{2} l + 2$.

By what is given, $\frac{5}{2} l + 2 = 42 \Rightarrow \frac{5}{2} l = 42 - 2 = 40 \Rightarrow l = 40 \cdot \frac{2}{5} = 16$

Hence, width$= w = \frac{1}{4} \cdot 16 + 1 = 5$

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 $\frac{x}{4} + 1$

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

$2 L + 2 W = 42$

sub in values and solve

$2 \left(x\right) + 2 \left(\left(\frac{x}{4}\right) + 1\right)$

$2 x + \frac{2 x}{4} + 2 = 42$

eliminate the fraction by multiplying everything by 4

$8 x + 2 x + 8 = 168$

collect like terms

$10 x = 160$

$\therefore L = 16$

sub in to find the width

$W = \left(\frac{16}{4}\right) + 1$

$W = 4 + 1$

$W = 5$

verify

$2 L + 2 W = 42$

$2 \left(16\right) + 2 \left(5\right) = 42$

$32 + 10 = 42$

$42 = 42$

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$\text{ } \to w = \frac{L}{4} + 1$ .................Equation(1)

The perimeter is 42 $\text{ } \to 2 w + 2 L = 42$............Equation(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From equation(2) $w = \frac{42}{2} - \frac{2}{2} L = 21 - L$ ....................Equation(3)

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

$\textcolor{b r o w n}{w = \frac{L}{4} + 1} \textcolor{b l u e}{\text{ "->" } 21 - L = \frac{L}{4} + 1}$

$- L - \frac{L}{4} = 1 - 21$

$- \frac{5}{4} L = - 20$

Multiply both sides by (-1)

$\frac{5}{4} L = 20$

$\textcolor{g r e e n}{\implies L = \frac{4}{5} \times 20 = 16}$

substitute for L in equation(2)

$2 w + 2 L = 42 \text{ "->" } 2 w + 2 \left(16\right) = 42$

$\textcolor{g r e e n}{w = 5}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
Using the wording of the question the following should be true:

$5 \to 1 + \left(\frac{1}{4} \times 16\right) = 5 \leftarrow \text{True}$