Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2005 AMC 10A Problems/Problem 8: Difference between revisions

← Older editNewer edit →
JWK750 (talk | contribs)
235 edits
No edit summary
Nathan wailes (talk | contribs)
2,960 edits
No edit summary
Line 23: Line 23:
[[Category:Introductory Geometry Problems]]
[[Category:Introductory Geometry Problems]]
[[Category:Area Problems]]
[[Category:Area Problems]]
{{MAA Notice}}

Revision as of 10:29, 4 July 2013

Problem

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE$=1. What is the area of the inner square $EFGH$?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42$

Solution

(C) We see that side $BE$, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, $AH = 1$. Then $HB = HE + BE = HE + 1$, and $HE$ is one of the sides of the square whose area we want to find. So:

$1^2 + (HE+1)^2=\sqrt{50}^2$

$1 + (HE+1)^2=50$

$(HE+1)^2=49$

$HE+1=7$

$HE=6$ So, the area of the square is $6^2=\boxed{36}$.

See Also

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&oldid=54266"