Difference between revisions of "2005 AMC 10A Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE</math> | + | In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE=1</math>. What is the area of the inner square <math>EFGH</math>? |
[[File:AMC102005Aq.png]] | [[File:AMC102005Aq.png]] | ||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I | + | We see that side <math>BE</math>, which we know is <math>1</math>, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So: |
<cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath> | <cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath> | ||
| Line 18: | Line 18: | ||
<cmath>HE=6</cmath> | <cmath>HE=6</cmath> | ||
| − | So, the area of the square is <math>6^2=\boxed{ | + | So, the area of the square is <math>6^2=\boxed{\textbf{(C) }36}</math> |
| − | ==See | + | ==See also== |
| + | {{AMC10 box|year=2005|ab=A|num-b=7|num-a=9}} | ||
| − | |||
| − | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:58, 13 December 2021
Problem
In the figure, the length of side 
of square 
is 
and 
. What is the area of the inner square 
?
Solution
We see that side 
, which we know is 
, is also the shorter leg of one of the four right triangles (which are congruent, I won’t prove this). So, 
. Then 
, and 
is one of the sides of the square whose area we want to find. So:
![\[1^2 + (HE+1)^2=\sqrt{50}^2\]](http://latex.artofproblemsolving.com/7/c/3/7c37e12e7b6f1641b19a232b4fdbdb6f03349aac.png)
![\[1 + (HE+1)^2=50\]](http://latex.artofproblemsolving.com/2/d/7/2d7e5f6ccce6d23fa35a57d269737c7e9b494cfb.png)
![\[(HE+1)^2=49\]](http://latex.artofproblemsolving.com/c/a/d/cad7941ecb53cad597279cc85a31fa4983b80f0b.png)
![\[HE+1=7\]](http://latex.artofproblemsolving.com/c/e/b/ceb2c8f74471c971c81fcf1364d33b5fbc98787b.png)
![\[HE=6\]](http://latex.artofproblemsolving.com/c/b/0/cb09cde4ae1381e2d11cd524c1aab06d502fb90e.png)
So, the area of the square is 
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
