Difference between revisions of "2023 AMC 10A Problems/Problem 11"
Ghfhgvghj10 (talk | contribs) |
(→Problem) |
||
Line 17: | Line 17: | ||
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(E--F--G--H--cycle); | draw(E--F--G--H--cycle); | ||
− | <asy> | + | </asy> |
<math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math> | <math>\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1</math> |
Revision as of 20:32, 9 November 2023
Problem
A square of area is inscribed in a square of area , creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
Solution
Note that each side length is and Let the shorter side of our triangle be , thus the longer leg is . Hence, by the Pythagorean Theorem, we have .
By the quadratic formula, we find . Hence, our answer is
~SirAppel ~ItsMeNoobieboy
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.