rectangle and square

• Nov 11th 2011, 02:52 AM
kingman
rectangle and square
I need help in the below problem

The figure is made up of a square and a rectangle. The ratio of the area of the square to the area of the rectangle is 2 : 5. I/4 of the square is shaded.
What fraction of the figure is unshaded?
• Nov 11th 2011, 03:07 AM
alexmahone
Re: rectangle and square
Quote:

Originally Posted by kingman
I need help in the below problem

The figure is made up of a square and a rectangle. The ratio of the area of the square to the area of the rectangle is 2 : 5. I/4 of the square is shaded.
What fraction of the figure is unshaded?

Let the areas of the square and the rectangle be 2 and 5 respectively.

Area of the shaded area = 1/4 * 2 = 0.5

Total area = 2 + 5 - 0.5 = 6.5

Area of the unshaded area = 6.5 - 0.5 = 6

Fraction unshaded = 6 / 6.5 = 12 / 13
• Nov 11th 2011, 06:11 AM
Soroban
Re: rectangle and square
Hello, kingman!

Another approach . . .

Quote:

The figure is made up of a square and a rectangle.
The ratio of the area of the square to the area of the rectangle is 2:5.
I/4 of the square is shaded.
What fraction of the figure is unshaded?

Code:

      * - - - - - - - *       |              |       |              |       |    C    * - + - *       |          |:::|  |       |          |:A:|  |       * - - - - - + - *  |                   |      |                   |  B  |                   * - - - *
Let $S$ = area of the square.
Let $R$ = area of the rectangle.

We have: . $A = \tfrac{1}{4}S,\;B = \tfrac{3}{4}S$

We have: . $\frac{S}{R} = \frac{2}{5} \quad\Rightarrow\quad R \,=\,\tfrac{5}{2}S$

Then: . $C \:=\:R - A \:=\:\tfrac{5}{2}S - \tfrac{1}{4}S \:=\:\tfrac{9}{4}S$

The unshaded area is: . $B+C \:=\:\tfrac{3}{4}S + \tfrac{9}{4}S \:=\:3S$

The total area is: . $A+B+C \:=\:\tfrac{1}{4}S + \tfrac{3}{4}S + \tfrac{9}{4}S \:=\:\tfrac{13}{4}S$

Therefore: . $\text{Fraction unshaded} \;=\; \frac{3S}{\frac{13}{4}S} \;=\;\frac{12}{13}$

• Nov 11th 2011, 08:36 AM
kingman
Re: rectangle and square
Quote:

Originally Posted by Soroban
Hello, kingman!

Another approach . . .

Code:

    * - - - - - - - *     |              |     |              |     |    C    * - + - *     |          |:::|  |     |          |:A:|  |     * - - - - - + - *  |                 |      |                 |  B  |                 * - - - *
Let $S$ = area of the square.
Let $R$ = area of the rectangle.

We have: . $A = \tfrac{1}{4}S,\;B = \tfrac{3}{4}S$

We have: . $\frac{S}{R} = \frac{2}{5} \quad\Rightarrow\quad R \,=\,\tfrac{5}{2}S$

Then: . $C \:=\:R - A \:=\:\tfrac{5}{2}S - \tfrac{1}{4}S \:=\:\tfrac{9}{4}S$

The unshaded area is: . $B+C \:=\:\tfrac{3}{4}S + \tfrac{9}{4}S \:=\:3S$

The total area is: . $A+B+C \:=\:\tfrac{1}{4}S + \tfrac{3}{4}S + \tfrac{9}{4}S \:=\:\tfrac{13}{4}S$

Therefore: . $\text{Fraction unshaded} \;=\; \frac{3S}{\frac{13}{4}S} \;=\;\frac{12}{13}$

I wonder why all the numbers and text displayed are blurred; so cannot figure out what is wriiten on the solution page.
thanks