SOLUTION
We are told to find the area of the green square.
Now the area of the red square is 16 square-feet
[tex]\begin{gathered} \text{area of square = }l^2 \\ 16=l^2 \\ \text{take square-root of both sides } \\ \sqrt[]{16}=\sqrt[]{l^2} \\ \text{square cancels square root } \\ 4=l \\ l=4ft \end{gathered}[/tex]
Hence one length of the red square which is b from the diagram is 4 ft.
And the area of the yellow square is 25 square-feet. This means
[tex]\begin{gathered} \text{area of square = }l^2 \\ 25=l^2 \\ \text{take square-root of both sides } \\ \sqrt[]{25}=\sqrt[]{l^2} \\ \text{square cancels square root } \\ 5=l \\ l=5ft \end{gathered}[/tex]
Hence one length of the yellow square which is c from the diagram is 5 ft.
Now, we have found two sides of the blue triangle. Let's see the sides
So we need to find the side a of the triangle. To do this, we will use the Pythagoras theorem. This says that the square of the longest side (the hypotenuse) equals the sum of the squares of the other two sides.
Using this we have
[tex]\begin{gathered} 5^2=a^2+4^2 \\ 25=a^2+16 \\ a^2=25-16 \\ a^2=9 \\ \sqrt[]{a^2}=\sqrt[]{9} \\ a=3\text{ ft} \end{gathered}[/tex]
Now, the third side a of the blue triangle is 3 feet. This side is one side of the green square.
Hence area of the green square becomes
[tex]\begin{gathered} \text{Area = }l^2 \\ \text{Area = 3}^2 \\ \text{Area = 9 ft } \end{gathered}[/tex]
The answer is therefore option B
