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Clara wanted to enlarge a figure on the photocopy machine. She set the enlargement factor to $1.25$. If the area of the figure was $10$ square inches before it was enlarged, what will the area be after it is run through the machine enlarged?

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  • $\begingroup$ Similar figures $\endgroup$ – Henry Nov 14 '14 at 19:38
  • $\begingroup$ Interesting question. Please, provide the context and doubts next time! $\endgroup$ – Quiet_waters Mar 7 '19 at 21:04
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Answer is 10 × 1.252 or 15.625.
How? The photocopy machine will increase the length of every side of the figure 1.25 times. So if there is square with side measuring 1cm. Its length will increase to 1.25cm. Now, you can see that its area will be 1.252 or 1.5625. Similarly, if there is a rectangle with length l and breadth b its sides will increase to 1.25 × l and 1.25 × b. So, the new area will be 1.25 × l × 1.25 × b or 1.252 × l × b or 1.252 × (OLD AREA).

This is because area is 2-dimensional. If this was the case with volume (maybe there was a 3-d printer instead), the figure's volume would have been increased by 1.253.

You an also try this with circles, polygons and other shapes. The results will be the same.

P.S.
Area of a circle is πr2 where π = 3.141 and r = radius of the circle

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  • $\begingroup$ So, enlargement factor always applies into the 'linear dimension' (sides in this case), I mean, how did you know that the enlargement factor did not apply to the area, that could becames directly $10\times 1.25$ ? Many thanks. $\endgroup$ – Quiet_waters Mar 4 '19 at 17:27
  • $\begingroup$ More specifically, if I say: enlarge the figure by an $100\%$ enlargement, can a people answer if we should multiply the area by $2$, instead of the lenghts? If not, there is some paper or statement about it? Many thanks. $\endgroup$ – Quiet_waters Mar 7 '19 at 21:07

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