Difference between revisions of "2012 AMC 8 Problems/Problem 25"
(→See Also) |
(→Solution 1) |
||
| Line 12: | Line 12: | ||
==Solution 1== | ==Solution 1== | ||
The total area of the four congruent triangles formed by the squares is <math>5-4 = 1 </math>. Therefore, the area of one of these triangles is <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>. | The total area of the four congruent triangles formed by the squares is <math>5-4 = 1 </math>. Therefore, the area of one of these triangles is <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>. | ||
| + | |||
| + | ==Solution 2 (Algebra)== | ||
==Video Solution 2== | ==Video Solution 2== | ||
Revision as of 21:42, 21 December 2024
Contents
Problem
A square with area 
is inscribed in a square with area 
, with each vertex of the smaller square on a side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length 
, and the other of length 
. What is the value of 
?
![[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]](http://latex.artofproblemsolving.com/3/d/4/3d4cdd3dbcabe3dc76b806d60b785416afc65f11.png)
Solution 1
The total area of the four congruent triangles formed by the squares is 
. Therefore, the area of one of these triangles is 
. The height of one of these triangles is and the base is . Using the formula for area of the triangle, we have 
. Multiply by 
on both sides to find that the value of is 
.
Solution 2 (Algebra)
Video Solution 2
https://youtu.be/MhxGq1sSA6U ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=2
~ pi_is_3.14
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by None | |
| 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. 
