Difference between revisions of "2005 AMC 12A Problems/Problem 7"
(solution) |
(→Solution 2) |
||
(7 intermediate revisions by 3 users not shown) | |||
Line 3: | Line 3: | ||
<math> | <math> | ||
− | (\mathrm {A}) \ 25 \qquad (\mathrm {B}) \ 32 \qquad (\mathrm {C})\ 36 \qquad (\mathrm {D}) \ | + | (\mathrm {A}) \ 25 \qquad (\mathrm {B}) \ 32 \qquad (\mathrm {C})\ 36 \qquad (\mathrm {D}) \ 40 \qquad (\mathrm {E})\ 42 |
</math> | </math> | ||
== Solution == | == Solution == | ||
+ | [[Image:2005_12A_AMC-7b.png]] | ||
+ | |||
Arguable the hardest part of this question is to visualize the diagram. Since each side of <math>EFGH</math> can be extended to pass through a vertex of <math>ABCD</math>, we realize that <math>EFGH</math> must be tilted in such a fashion. Let a side of <math>EFGH</math> be <math>x</math>. | Arguable the hardest part of this question is to visualize the diagram. Since each side of <math>EFGH</math> can be extended to pass through a vertex of <math>ABCD</math>, we realize that <math>EFGH</math> must be tilted in such a fashion. Let a side of <math>EFGH</math> be <math>x</math>. | ||
Line 11: | Line 13: | ||
Notice the [[right triangle]] (in blue) with legs <math>1, x+1</math> and [[hypotenuse]] <math>\sqrt{50}</math>. By the [[Pythagorean Theorem]], we have <math>1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6</math>. Thus, <math>[EFGH] = x^2 = 36\ \mathrm{(C)}</math> | Notice the [[right triangle]] (in blue) with legs <math>1, x+1</math> and [[hypotenuse]] <math>\sqrt{50}</math>. By the [[Pythagorean Theorem]], we have <math>1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6</math>. Thus, <math>[EFGH] = x^2 = 36\ \mathrm{(C)}</math> | ||
− | |||
== See also == | == See also == | ||
{{AMC12 box|year=2005|num-b=6|num-a=8|ab=A}} | {{AMC12 box|year=2005|num-b=6|num-a=8|ab=A}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 02:49, 5 November 2022
Problem
Square is inside the square
so that each side of
can be extended to pass through a vertex of
. Square
has side length
and
. What is the area of the inner square
?
Solution
Arguable the hardest part of this question is to visualize the diagram. Since each side of can be extended to pass through a vertex of
, we realize that
must be tilted in such a fashion. Let a side of
be
.
Notice the right triangle (in blue) with legs and hypotenuse
. By the Pythagorean Theorem, we have
. Thus,
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.