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ollegr [7]
2 years ago
12

He dimensions are tripled. The new surface area would be times larger than the original surface area.

Mathematics
1 answer:
ddd [48]2 years ago
3 0

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.

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Answer:

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Step-by-step explanation:

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thus, for ∆ABC\cong∆MTR

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thus,for ∆ABC\cong∆MTR

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likewise,

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