Answer:
The new surface area would be 9 times larger than the original surface area.
Step-by-step explanation:
Here we do not know what was the original shape, but we will see that it does not matter.
Let's start with a square of side length L.
The original area of this square will be:
A = L^2
Now if each dimension is tripled, then all the sides of the square now will be equal to 3*L
Then the new area of the square is:
A' = (3*L)^2 = (3*L)*(3*L) = 9*L^2 = 9*A
So the new surface area is 9 times the original one.
Now, if the figure was a circle instead of a square?
For a circle of radius R, the area is:
A = pi*R^2
where pi = 3.14
Now if the dimensions of the circle are tripled, the new radius will be 3*R
Then the new area of the circle is:
A' = pi*(3*R)^2 = pi*9*R^2 = 9*(pi*R^2) = 9*A
Again, the new area is 9 times the original one.
If the figure is a triangle?
We know that for a triangle of base B and height H, the area is:
A = B*H/2
If we triple each measure, we will have a base 3*B and a height 3*H
Then the new area is:
A' = (3*B)*(3*H)/2 = (3*3)*(B*H/2) = 9*(B*H/2) = 9*A
Again, the new area is 9 times the original area.
So we can conclude that for any shape, the new area will be 9 times the original area.
