Difference between revisions of "2010 AMC 10A Problems/Problem 2"

(Solution 2)
m (Solution 3)
Line 35: Line 35:
  
 
==Solution 3==
 
==Solution 3==
Let the side length of one of the squares equal <math>1</math>. Then, the width of the rectangle will be <math>4</math>, and since the width of the rectangle is the same as the length of the entire shape, the length of the rectangle is <math>4 - 1 = 3</math>. The ratio between the two is therefore <math>\frac{4}{3}</math>, so our answer is <math>B</math>.  
+
Let the side length of one of the squares equal <math>1</math>. Then, the width of the rectangle will be <math>4</math>, and since the width of the rectangle is the same as the length of the entire shape, the length of the rectangle is <math>4 - 1 = 3</math>. The ratio between the two is therefore <math>\frac{4}{3}</math>, so our answer is <math>\boxed{B}</math>.  
  
 
~ abed_nadir
 
~ abed_nadir

Revision as of 12:12, 11 November 2023

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?

[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]

$\mathrm{(A)}\ \dfrac{5}{4} \qquad \mathrm{(B)}\ \dfrac{4}{3} \qquad \mathrm{(C)}\ \dfrac{3}{2} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 3$

Solution 1

Let the length of the small square be $x$, intuitively, the length of the big square is $4x$. It can be seen that the width of the rectangle is $3x$. Thus, the length of the rectangle is $4x/3x = 4/3$ times as large as the width. The answer is $\boxed{B}$.

Solution 2

We can say the area of one small square is $x^2$, so $\dfrac{1}{4}$ of the area of the large square is $4x^2$ so the area of the large square is $16x^2$, so each side is $4x$ so the length of the rectangle is $4x$ and the width of the rectangle is $4x-x=3x$ so $\dfrac{4x}{3x}=\dfrac{4}{3}$

Solution 3

Let the side length of one of the squares equal $1$. Then, the width of the rectangle will be $4$, and since the width of the rectangle is the same as the length of the entire shape, the length of the rectangle is $4 - 1 = 3$. The ratio between the two is therefore $\frac{4}{3}$, so our answer is $\boxed{B}$.

~ abed_nadir

Video Solution

https://youtu.be/C1VCk_9A2KE?t=80

~IceMatrix

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png