rectangle and square

**kingman** · November 11th, 2011, 02:52 AM

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?

**alexmahone** · November 11th, 2011, 03:07 AM

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

**Soroban** · November 11th, 2011, 06:11 AM

Hello, kingman!

Another approach . . .

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

**kingman** · November 11th, 2011, 08:36 AM

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

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