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?
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?
Let $\displaystyle S$ = area of the square.Code:* - - - - - - - * | | | | | C * - + - * | |:::| | | |:A:| | * - - - - - + - * | | | | B | * - - - *
Let $\displaystyle R$ = area of the rectangle.
We have: .$\displaystyle A = \tfrac{1}{4}S,\;B = \tfrac{3}{4}S$
We have: .$\displaystyle \frac{S}{R} = \frac{2}{5} \quad\Rightarrow\quad R \,=\,\tfrac{5}{2}S$
Then: .$\displaystyle C \:=\:R - A \:=\:\tfrac{5}{2}S - \tfrac{1}{4}S \:=\:\tfrac{9}{4}S$
The unshaded area is: .$\displaystyle B+C \:=\:\tfrac{3}{4}S + \tfrac{9}{4}S \:=\:3S$
The total area is: .$\displaystyle A+B+C \:=\:\tfrac{1}{4}S + \tfrac{3}{4}S + \tfrac{9}{4}S \:=\:\tfrac{13}{4}S$
Therefore: .$\displaystyle \text{Fraction unshaded} \;=\; \frac{3S}{\frac{13}{4}S} \;=\;\frac{12}{13} $