# Rectangle problem-probably easy

• Oct 29th 2008, 02:39 PM
alaricepent
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
• Oct 29th 2008, 04:00 PM
Soroban
Hello, alaricepent!

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

Quote:

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)