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On an aerial photograph, the surface of a pond appears as

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Joined: 22 Jun 2010
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On an aerial photograph, the surface of a pond appears as  [#permalink]

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Updated on: 20 May 2012, 11:17
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67% (02:03) correct 33% (02:03) wrong based on 266 sessions

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On an aerial photograph, the surface of a pond appears as circular region of radius 7/16 inch. If a distance of 1 inch on the photograph corresponds to an actual distance of 2 miles, which of the following is the closest estimate of the actual surface area of the pond, in square miles?

A. 1.3
B. 2.4
C. 3.0
D. 3.8
E. 5.0

Originally posted by mehdiov on 07 Sep 2010, 14:13.
Last edited by Bunuel on 20 May 2012, 11:17, edited 1 time in total.
Edited the question and added the OA
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07 Sep 2010, 14:31
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mehdiov please post the questions in their original form and provide answer choice with them.

On an aerial photograph, the surface of a pond appears as circular region of radius 7/16 inch. If a distance of 1 inch on the photograph corresponds to an actual distance of 2 miles, which of the following is the closest estimate of the actual surface area of the pond, in square miles?

A. 1.3
B. 2.4
C. 3.0
D. 3.8
E. 5.0

As 1 inch = 2 miles, then 7/16 inches = 2*7/16 = 7/8 miles. So actual radius is 7/8 miles. $$Area=\pi{r^2}\approx{\frac{22}{7}*\frac{49}{64}}\approx{2.4}$$ (as we are asked about the approximate value we can take $$\pi\approx{\frac{22}{7}}$$).

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Joined: 22 Jun 2010
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07 Sep 2010, 14:35
Bunuel wrote:
mehdiov please post the questions in their original form and provide answer choice with them.

On an aerial photograph, the surface of a pond appears as circular region of radius 7/16 inch. If a distance of 1 inch on the photograph corresponds to an actual distance of 2 miles, which of the following is the closest estimate of the actual surface area of the pond, in square miles?

A. 1.3
B. 2.4
C. 3.0
D. 3.8
E. 5.0

As 1 inch = 2 miles, then 7/16 inches = 2*7/16 = 7/8 miles. So actual radius is 7/8 miles. $$Area=\pi{r^2}\approx{\frac{3.14*49}{64}}\approx{2.4}$$.

thanks. appoligies for that
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Re: On an aerial photograph, the surface of a pond appears as  [#permalink]

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03 Aug 2017, 11:42
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Why doesn't converting the area to square inches and then converting from inches to miles work?

$$(\frac{7}{16})^2=\frac{49}{256}$$
$$\frac{49}{256}*\frac{22}{7}=\frac{77}{128}$$
$$\frac{77}{12}*2\approx1.2$$

I chose 1.3 because it's the closest answer, but obviously the conversion to miles needs to happen before calculating the area. Why does this matter, and is there a decent framework to do conversions of this sort that involve squaring and cubing? This seems to pop up often, and I find it slightly confusing. Thanks!

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Re: On an aerial photograph, the surface of a pond appears as  [#permalink]

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30 Aug 2017, 04:55
I chose 1.3miles as well but I think they key word here is "square miles". That's the indicator that you need to convert before using the area formula
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On an aerial photograph, the surface of a pond appears as  [#permalink]

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31 Aug 2017, 12:17
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spence11 wrote:
Why doesn't converting the area to square inches and then converting from inches to miles work?

$$(\frac{7}{16})^2=\frac{49}{256}$$
$$\frac{49}{256}*\frac{22}{7}=\frac{77}{128}$$
$$\frac{77}{12}*2\approx1.2$$

I chose 1.3 because it's the closest answer, but obviously the conversion to miles needs to happen before calculating the area. Why does this matter, and is there a decent framework to do conversions of this sort that involve squaring and cubing? This seems to pop up often, and I find it slightly confusing. Thanks!

SR

spence11 , these problems can be confusing. We have a "scale factor" issue, as you note.

In one sense, however, when scaling up or down, linear change is one kind of step (ONE change in one length), and area change is a different kind of step (TWO changes, one for each length).

"...converting finding the area to in square inches and then converting from inches to miles [does not] work" because you are not converting from inches to miles.

You need to convert square inches (in the area you found for the pond on the photo) to square miles.

The scale factor here is
(2 miles: 1 inch) = 2

When converting an area, you have to use the scale factor, $$k$$, twice.

Area conversion requires $$k^2$$ because
-- the area of any two dimensional figure, circles included, is (length times length)

To convert area using the change from one linear unit to another (distance in inches -> distance in miles) ...
what you do to one length (scale it up), you must do to the other.

And to calculate area, we use two lengths. You converted only one of the two lengths.

You chose the correct scale factor of 2. But you need to multiply the area you found by $$k^2=2^2=4$$.
You are short one factor of $$k$$. Close!

($$\frac{77}{128}$$ * 4) = $$\frac{77}{32}$$ = 2.4 (sq mi - see below)

Here is one way to look at it:

$$\frac{1in}{2mi}$$ -- those are one-dimensional lengths. Lines.

Convert them to two-dimensional areas by squaring the number and the unit:

(1 in)$$^2$$ = (1 in)*(1 in) = 1 sq. inch, or 1 in$$^2$$

(2 mi)$$^2$$ = (2 mi)(2 mi) = 4 square miles, or 4mi$$^2$$

Below, area in square inches is converted to area in square miles.
Square inches cancel, leaving square miles as the desired result, but only because I multiply the original area by the scale factor twice:

$$\frac{77}{128}(in*in)$$ * $$\frac{2mi}{1in}$$ * $$\frac{2mi}{1in}$$ = $$\frac{77}{32}$$(mi*mi) =

2.4 miles$$^2$$

So for any conversion from area to area: if you find the original area in small units, multiply that area by the scale factor squared to get the larger actual area.

Bunuel sidestepped the whole issue by
-- multiplying one length (radius in inches) by one scale factor (2) to get [another] one length: the radius in miles.
Then he found the area in square miles with normal calculations for a circle's area.

Length: multiply by one scale factor $$k$$

Area = length * length, so multiply by two scale factors $$k^2$$

Volume: multiply by three scale factors $$k^3$$ (b/c V = L * L * L)

It depends on the problem, but often it is easier to convert one length of the smaller figure to one length of the larger figure (here, radius), and then calculate area or volume.
If you do it that way, you only deal with the scale factor once: length --> length.

Hope that helps.
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Re: On an aerial photograph, the surface of a pond appears as  [#permalink]

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02 Sep 2017, 06:09
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mehdiov wrote:
On an aerial photograph, the surface of a pond appears as circular region of radius 7/16 inch. If a distance of 1 inch on the photograph corresponds to an actual distance of 2 miles, which of the following is the closest estimate of the actual surface area of the pond, in square miles?

A. 1.3
B. 2.4
C. 3.0
D. 3.8
E. 5.0

We can create a proportion in which n is the radius of the pond in miles:

1/2 = (7/16)/n

n = 2(7/16)

n = 7/8

Thus, the area of the pond is (7/8)^2 x π. Recall that the fractional approximation of π is 22/7. Thus, we have:

(7/8)^2 x π ≈ 49/64 x 22/7 = 7/32 x 11/1 = 77/32 = 2 13/32 ≈ 2 12/30 = 2 2/5 = 2.4

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Re: On an aerial photograph, the surface of a pond appears as  [#permalink]

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07 Sep 2017, 04:56
Did a very silly calculation mistake -> 8*4 = 64
How to avoid such careless mistakes?
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Re: On an aerial photograph, the surface of a pond appears as  [#permalink]

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15 Mar 2020, 07:09
first convert 7/16 inch into miles by multiplying with 2 then find surface are of pond using formula pie r^2
Re: On an aerial photograph, the surface of a pond appears as   [#permalink] 15 Mar 2020, 07:09