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> | + | In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math>, <math>E</math> is between <math>B</math> and <math>H</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)); | ||
| − | <math> \textbf{(A)} | + | 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> | ||
| + | \textbf{(A) } 25\qquad\textbf{(B) } 32\qquad\textbf{(C) } 36\qquad\textbf{(D) } 40\qquad\textbf{(E) } 42 | ||
| + | </math> | ||
| + | |||
| + | ==Solution== | ||
| + | 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 obviously congruent, using the symmetry of the diagram). So <math>AH = 1</math>, and hence <math>HB = HE + BE = HE + 1</math>. Since <math>HE</math> is one of the sides of the square whose area we want to find, we can now simply apply Pythagoras' theorem: | ||
| + | |||
| + | <cmath>\begin{align*}1^2 + (HE+1)^2 = \left(\sqrt{50}\right)^2 &\iff 1 + (HE+1)^2 = 50 \\ &\iff (HE+1)^2 = 49 \\&\iff HE+1 = 7 \qquad \text{(as } HE \text{ is a length, so must be positive)} \\ &\iff HE = 6.\end{align*}</cmath> | ||
| + | |||
| + | Thus the area of the square is <math>6^2 = \boxed{\textbf{(C) }36}</math>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2005|ab=A|num-b=7|num-a=9}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 17:09, 1 July 2025
Problem
In the figure, the length of side 
of square 
is 
, 
is between 
and 
, and 
. What is the area of the inner square 
?
![[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]](http://latex.artofproblemsolving.com/6/3/5/635040d9dfc222611316ef10bd688c50783d97b7.png)
Solution
We see that side 
, which we know is 
, is also the shorter leg of one of the four right triangles (which are obviously congruent, using the symmetry of the diagram). So 
, and hence 
. Since 
is one of the sides of the square whose area we want to find, we can now simply apply Pythagoras' theorem:
Thus 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
