Difference between revisions of "2006 AMC 8 Problems/Problem 5"
m (→Problem) |
|||
| (5 intermediate revisions by 4 users not shown) | |||
| Line 16: | Line 16: | ||
\textbf{(D)}\ 30 \qquad | \textbf{(D)}\ 30 \qquad | ||
\textbf{(E)}\ 40 </math> | \textbf{(E)}\ 40 </math> | ||
| − | |||
== Solution == | == Solution == | ||
| Line 26: | Line 25: | ||
Thus, the area of the smaller square in the picture is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | Thus, the area of the smaller square in the picture is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | ||
| + | |||
| + | ===Solution 3=== | ||
| + | The smaller square forms four congruent isosceles right triangles whose legs measure half of the larger square's side length. Since the larger square has an area of <math>60</math>, its sides measure <math>\sqrt{60}=2\sqrt{15}</math>. Therefore, the legs of the right triangles measure <math>\sqrt{15}</math>. Now we can use the Pythagorean Theorem to find the length of the hypotenuses, or the smaller square's sides. We have | ||
| + | <cmath>\sqrt{{\sqrt{15}}^2+{\sqrt{15}}^2}=\sqrt{15+15}=\sqrt{30}.</cmath> | ||
| + | Squaring this, the area of the smaller square is <math> \boxed{\textbf{(D)}\ 30 } </math>. | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/j3QSD5eDpzU?t=822 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/pfqtAX6AHNg | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=4|num-a=6}} | {{AMC8 box|year=2006|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:20, 29 October 2024
Contents
Problem
Points 
and 
are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
![[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]](http://latex.artofproblemsolving.com/e/7/8/e7899c538a174faf9963a60d09886b7fb12ae942.png)
Solution
Solution 1
Drawing segments 
and 
, the number of triangles outside square 
is the same as the number of triangles inside the square. Thus areas must be equal so the area of is half the area of the larger square which is 
.
Solution 2
If the side length of the larger square is 
, the side length of the smaller square is 
. Therefore the area of the smaller square is 
, half of the larger square's area, 
.
Thus, the area of the smaller square in the picture is .
Solution 3
The smaller square forms four congruent isosceles right triangles whose legs measure half of the larger square's side length. Since the larger square has an area of 
, its sides measure 
. Therefore, the legs of the right triangles measure 
. Now we can use the Pythagorean Theorem to find the length of the hypotenuses, or the smaller square's sides. We have
![\[\sqrt{{\sqrt{15}}^2+{\sqrt{15}}^2}=\sqrt{15+15}=\sqrt{30}.\]](http://latex.artofproblemsolving.com/0/0/a/00a0c00a3fddc8bbf26eba462d64744c0a5942b1.png)
Squaring this, the area of the smaller square is 
.
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=822
~ pi_is_3.14
Video Solution by WhyMath
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
