Difference between revisions of "2008 AMC 10A Problems/Problem 2"
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==Solution 1== | ==Solution 1== | ||
| − | Since they are asking for the "ratio" of two things, we can say that the side of the square is anything that we want. So if we say that it is <math>1</math>, then width of the rectangle is <math>2</math>, and the length is <math>4</math>, thus making the total area of the rectangle <math>8</math>. The area of the square is just <math>1</math>. So the | + | Since they are asking for the "ratio" of two things, we can say that the side of the square is anything that we want. So if we say that it is <math>1</math>, then the width of the rectangle is <math>2</math>, and the length is <math>4</math>, thus making the total area of the rectangle <math>8</math>. The area of the square is just <math>1</math>. So the percent of the rectangle's area inside the square is <math>\frac{1}{8} \times 100 = 12.5 \longrightarrow \fbox{A}</math> |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2008|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:20, 4 June 2021
Problem
A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is 
. The ratio of the rectangle's length to its width is . What percent of the rectangle's area is inside the square?
Solution 1
Since they are asking for the "ratio" of two things, we can say that the side of the square is anything that we want. So if we say that it is 
, then the width of the rectangle is 
, and the length is 
, thus making the total area of the rectangle 
. The area of the square is just . So the percent of the rectangle's area inside the square is 
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
