2005 AMC 12A Problems/Problem 7: Difference between revisions
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== Problem == | == Problem == | ||
[[Square]] <math>EFGH</math> is inside the square <math>ABCD</math> so that each side of <math>EFGH</math> can be extended to pass through a vertex of <math>ABCD</math>. Square <math>ABCD</math> has side length <math>\sqrt {50}</math> and <math>BE = 1</math>. What is the area of the inner square <math>EFGH</math>? | [[Square]] <math>EFGH</math> is inside the square <math>ABCD</math> so that each side of <math>EFGH</math> can be extended to pass through a vertex of <math>ABCD</math>. Square <math>ABCD</math> has side length <math>\sqrt {50}</math> and <math>BE = 1</math>. What is the area of the inner square <math>EFGH</math>? | ||
<asy> | |||
unitsize(4cm); | |||
defaultpen(linewidth(.8pt)+fontsize(10pt)); | |||
pair D=(0,0), C=(1,0), B=(1,1), A=(0,1); | |||
pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0]; | |||
pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H); | |||
draw(A--B--C--D--cycle); | |||
draw(D--F); | |||
draw(C--E); | |||
draw(B--H); | |||
draw(A--G); | |||
label("$A$",A,NW); | |||
label("$B$",B,NE); | |||
label("$C$",C,SE); | |||
label("$D$",D,SW); | |||
label("$E$",E,NNW); | |||
label("$F$",F,ENE); | |||
label("$G$",G,SSE); | |||
label("$H$",H,WSW); | |||
</asy> | |||
<math> | <math> | ||
(\mathrm {A}) \ 25 \qquad (\mathrm {B}) \ 32 \qquad (\mathrm {C})\ 36 \qquad (\mathrm {D}) \ 40 \qquad (\mathrm {E})\ 42 | (\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]] | [[Image:2005_12A_AMC-7b.png]] | ||
Revision as of 10:15, 19 July 2024
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 | |
These problems are copyrighted © by the Mathematical Association of America.

