Difference between revisions of "2005 AMC 10A Problems/Problem 20"
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<math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 </math> | <math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7 </math> | ||
− | ==Solution== | + | ==Solution 1== |
The area of the octagon can be divided up into 5 squares with side <math>\frac{\sqrt2}2</math> and 4 right triangles, which are half the area of each of the squares. | The area of the octagon can be divided up into 5 squares with side <math>\frac{\sqrt2}2</math> and 4 right triangles, which are half the area of each of the squares. | ||
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<asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy> | <asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy> | ||
+ | ==Solution 2== | ||
+ | Using the diagram from above, we can extend the sides of length one to form four right triangles and the octagon, all inside a square. The right triangles are 45-45-90 triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>, so the side length is <math>\frac{1}{2}</math>. Thus, the area of the larger square is <math>2 \cdot 2 = 4</math>, and the area of the four right triangles combined is <math>\frac{1}{2}</math>, so the area of the octagon is <math>\frac{7}{2}</math>, or <math> \boxed{\mathrm{(A). \ }}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 23:30, 20 October 2017
Contents
Problem
An equiangular octagon has four sides of length 1 and four sides of length , arranged so that no two consecutive sides have the same length. What is the area of the octagon?
Solution 1
The area of the octagon can be divided up into 5 squares with side and 4 right triangles, which are half the area of each of the squares.
Therefore, the area of the octagon is equal to the area of squares.
The area of each square is , so the area of 7 squares is .
Solution 2
Using the diagram from above, we can extend the sides of length one to form four right triangles and the octagon, all inside a square. The right triangles are 45-45-90 triangles with hypotenuse , so the side length is . Thus, the area of the larger square is , and the area of the four right triangles combined is , so the area of the octagon is , or
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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