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namstech, Electrical Engineer

Category: Math Homework

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Experience: M.S. Signals and Information Processing

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Choose one of the following scenarios, explain your reasoning

Customer Question

Choose one of the following scenarios, explain your reasoning and provide an example that demonstrates your logic. Ø What happens to the circumference of a circle if you double the radius? What happens if you double the diameter? What happens if you triple the radius? Ø What happens to the area of a circle if you double the radius? What happens if you double the diameter? What happens if you triple the radius? Ø What is the effect on the area of a triangle if the base is doubled and the height is cut in half? What happens to the area, if the base is doubled and the height remains the same?

Q) What happens to the circumference of a circle if you double the radius? What happens if you double the diameter? What happens if you triple the radius?

SOLUTION:

Circumference = 2πr

- Let r1 be a radius, then circumference c1 is given as

c1 = 2πr1

If r2 = 2r1 (radius doubled), then

c2 = 2πr2 = 2π(2r1) = 2(2πr1) = 2c1

Hence when radius is doubled, circumference is doubled.

- In terms of diameter d, circumference is given as:

circumference = πd

For diameter d1, circumference c1 is given as

c1 = πd1

If d2 = 2d1, then

c2 = πd2 = π(2d1) = 2πd1 = 2c1

Hence when the diameter is doubled, circumference doubles.

If r3 = 3r1, then

c3 = 2πr3 = 2π(3r1) = 3 (2πr1) = 3c1

Hence, when radius is tripled, circumference triples.

Q) What happens to the area of a circle if you double the radius? What happens if you double the diameter? What happens if you triple the radius?

Hence, when the diameter is doubled, area is quadrupled.

Now, if r3 = 3r1, then

A3 = π(r3)^2 = π(3r1)^2 = 9 π(r1)^2 = 9A1

Hence, when the radius is tripled, area increases 9 times.

Q) What is the effect on the area of a triangle if the base is doubled and the height is cut in half? What happens to the area, if the base is doubled and the height remains the same?

If r2 = 2r1, then A2 = π (r2)^2 = π (2r1)^2 = 4 π (r1)^2 = 4 A1

Here, new radius r2 is double than that of radius r1. Now when you put that value in the formula, we get square of the value 2r1, whose square is 4r1^2 (i.e. square of 2 and square of r1^2). That is how we get 4 in the formula. Same argument goes for the radius r3 that is 3 times r1 i.e. 3r1^2. Squaring it in the formula yields the value 9r1^2 and hence we have 9 in the area.

I hope this helps. If there is still some confusion, please don't hesitate to ask.

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