Difference between revisions of "2010 AMC 10A Problems/Problem 2"
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− | 4/3 | + | == Problem 2 == |
+ | |||
+ | Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? | ||
+ | |||
+ | <center><asy> | ||
+ | unitsize(8mm); | ||
+ | defaultpen(linewidth(.8pt)); | ||
+ | |||
+ | draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); | ||
+ | draw((0,3)--(0,4)--(1,4)--(1,3)--cycle); | ||
+ | draw((1,3)--(1,4)--(2,4)--(2,3)--cycle); | ||
+ | draw((2,3)--(2,4)--(3,4)--(3,3)--cycle); | ||
+ | draw((3,3)--(3,4)--(4,4)--(4,3)--cycle); | ||
+ | |||
+ | </asy></center> | ||
+ | |||
+ | <math> | ||
+ | \mathrm{(A)}\ \dfrac{5}{4} | ||
+ | \qquad | ||
+ | \mathrm{(B)}\ \dfrac{4}{3} | ||
+ | \qquad | ||
+ | \mathrm{(C)}\ \dfrac{3}{2} | ||
+ | \qquad | ||
+ | \mathrm{(D)}\ 2 | ||
+ | \qquad | ||
+ | \mathrm{(E)}\ 3 | ||
+ | </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Let the length of the small square be <math>x</math>, intuitively, the length of the big square is <math>4x</math>. It can be seen that the width of the rectangle is <math>3x</math>. Thus, the length of the rectangle is <math>4x/3x = 4/3</math> times large as the width. The answer is <math>\boxed{B}</math>. |
Revision as of 15:35, 20 December 2010
Problem 2
Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?
Solution
Let the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is . Thus, the length of the rectangle is times large as the width. The answer is .