1. ## Rectangle problem-probably easy

Suppose the length of a rectangle is twice the width. If we increase both the width and length of the rectangle by 2 units, the area is increased by 34 square units. Find the width of the rectangle.

Can someone explain the process of finding the answer?

a)3
b)8
c)10
d)5
e)none of these

2. Hello, alaricepent!

Make a sketch, then baby-talk your way through it . . .

Suppose the length of a rectangle is twice the width.
If we increase both the width and length of the rectangle by 2 units,
the area is increased by 34 square units.
Find the width of the rectangle.

. . $\displaystyle (a)\;3 \qquad (b)\;8 \qquad (c)\;10 \qquad (d)\;5 \qquad(e)\text{ none of these}$
Code:
            2x
*-------------*
|             |
x |             | x
|             |
*-------------*
2x
Let $\displaystyle x$ = width
Then $\displaystyle 2x$ = length.
. . The area is: .$\displaystyle A_1 \:=\:(2x)(x) \:=\:2x^2$

Code:
             2x+2
*-------------*---*
|             :   |
|             :   |
x+2  |             :   | x+2
* - - - - - - *   |
|                 |
*-----------------*
2x+2
Now $\displaystyle x+2$ = width
and $\displaystyle 2x+2$ = length
. . The area is: .$\displaystyle A_2 \;=\;(x+2)(2x+2) \:=\:2x^2+6x+4$

The new area is 34 more than the original area:
. . $\displaystyle 2x^2 + 6x + 4 \;=\;2x^2 + 34$

So we have: .$\displaystyle 6x \:=\:30 \quad\Rightarrow\quad x \:=\:5$

The width is 5 . . . answer (d)