Question

Given this figure, if you increase the height by a factor of 3, how much larger will the area be?

halved

doubled

increased by two units

tripled

Answer 1

What is the formula for the area of a triangle?

Answer 2

\[a=\frac{ 1 }{ 2 }*b*h \]

Answer 3

Right. So if you substitute \(3h\) for \(h\), what happens to \(a\)?

Answer 4

it cancels out?

Answer 5

No...

if we first had \[a = \frac{1}{2}b*h\]and we replace \(h\) with \(3h\) we get
\[A = \frac{1}{2}b*(3h) = \frac{3}{2}bh\]Which of the answer choices does that represent?

Answer 6

compare \[\frac{3}2 bh\]with \[\frac{1}{2}bh\]

Answer 7

tripled

Answer 8

So, these problems where you have some figure that is doubled in length, halved in width, etc. Just plug the changes (doubled = 2, halved = 1/2, etc.) into the formula and you'll get the ratio out:

For this one, we have \[A = \frac{1}{2}bh\] We left the base alone, so we put in a 1 for \(b\), we tripled the height, so we put in a 3 for \(h\), result is
\[b*h = 3\] (the other factors we can ignore, as they are identical for both)

Answer 9

If we had a sphere whose volume is given by \[V =\frac{4}{3}\pi r^3\] and we double the radius, we put in \(r = 2\) and find that the volume has gone up by a factor of 8, because \[r^3 = (2)^3 = 8\]

Answer 10

Does that make sense, or have I just confused you? :-)

Answer 11

confused me

Answer 12

Okay, would you prefer to forget what I said, or should I try it in a bit more detail, with smaller steps?

Answer 13

I'll try again, you can read or ignore at your choice.

We had a triangle. Let's say that the base of the triangle was length \(b_0\) and the height of the triangle was length \(h_0\). We know that the formula for the area of a triangle is \[A = \frac{1}{2}bh\] where \(b\) is the length of the base, and \(h\) is the height.

For our initial triangle, the area was
\[A_0=\frac{1}{2}b_0*h_0\]

Our second triangle (bigger, improved!) had the height tripled to \(3*h_0\)
giving the new area as
\[A_{1} = \frac{1}{2}b_0*(3*h_0) = \frac{3}{2}b_0h_0\]

The ratio of the new triangle's area to the old triangle's area is
\[\frac{A_1}{A_0} = \frac{\frac{3}{2}b_0h_0}{\frac{1}{2}b_0h_0} = 3\]

Agreed?

Answer 14

What if besides tripling the height, we had doubled the base?

\[A_2 = \frac{1}{2}(2*b_0)(3*h_0) = 3b_0h_0\]

Comparing that with the original
\[\frac{A_2}{A_0} = \frac{3b_0h_0}{\frac{1}{2}b_0h_0} = 6\]

Now let's try the area of a circle with radius \(r\), given by \(A = \pi r^2\)

\[A_0 = \pi r_0^2\]
We double the radius of the circle to \(2r_0\)
\[A_1 = \pi (2 *r_0)^2 = 4\pi r_0^2\]
Ratio of new to old =\[\frac{A_1}{A_0} = \frac{4\pi r_0^2}{\pi r_0^2} = 4\]

In each of these cases, the ratio is just the result of the ratios of new to old being put in place of the variables. For the triangle, the formula is a constant * b*h, so if we double the base and triple the height, the ratio is just 2*3 = 6
For the circle, the formula is a constant * r^2, so if we double the radius, the ratio is (2)^2 = 4.

Any less confusing?

Answer 15

i would really just like to know the answer :/

Answer 16

so its doubled?

Answer 17

We got the answer, long ago. The ratio of the area of the new triangle to that of the old was 3, was it not?

Answer 18

A) halved
B)doubled
C)increased by two units
D)tripled

Answer 19

The ratio of the new to the old is 3. If you first have 10 dollars, and then you have 30 dollars, have you

a) halved the number of dollars
b) doubled the number of dollars
c) increased the number of dollars by two
d) tripled the number of dollars
?

Answer 20

halved = multiplied by 1/2
doubled = multiplied by 2
increased by 2 = added 2
tripled = multiplied by 3