Complete question:
Marvin is painting three squares shaped like the one below if it takes him exactly 4 cans of paint to paint square 3, five cans of paint to paint square 2 to how many cans of paint should he need to paint square one
Picture of squares is attached.
Answer:
9 cans of paint.
To paint square 1, he needs 9 cans of paint.
Step-by-step explanation:
Given that to paint square 3, four cans of paint are needed.
Therefore, let's take area of the square 3 as 4.
i.e a² = 4
Solving further, we have :
[tex] a = \sqrt{4} [/tex]
a = 2
We can say, each side of square 3 = 2 units
To paint square 2, five cans of paint are needed. Let's take the area of 2 to be 5.
i.e b² = 5
Solving further, we have :
[tex] b = \sqrt{5} [/tex]
We can say, each side of square 2 = [tex] \sqrt{5} [/tex] units
Using Pythagoras theorem, we have that the area of the square of a hypotenuse in a 90° angle is equal to the sum of the areas of the other two squares.
Therefore,
[tex] h^2 = 2^2 + \sqrt{5}^2 [/tex]
h² = 4 + 5
[tex] h = \sqrt{9} [/tex]
h = 3 units.
We now know one side of the square 1 is 3 units.
Area of square 1 = 3²
Area = 9.
Therefore to paint square 1, he needs 9 cans of paint.
