Difference between revisions of "2021 AMC 12A Problems/Problem 17"
Sugar rush (talk | contribs) |
|||
| Line 4: | Line 4: | ||
<math>\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215</math> | <math>\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215</math> | ||
| − | ==Solution== | + | == Video Solution (Using Similar Triangles, Pythagorean Theorem) == |
| − | + | https://youtu.be/gjeSGJy_ld4 | |
| + | |||
| + | ~ pi_is_3.14 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2021|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:29, 11 February 2021
Problem
Trapezoid 
has 
, and 
. Let 
be the intersection of the diagonals 
and 
, and let 
be the midpoint of . Given that 
, the length of 
can be written in the form 
, where 
and 
are positive integers and is not divisible by the quare of any prime. What is 
?
Video Solution (Using Similar Triangles, Pythagorean Theorem)
~ pi_is_3.14
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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. 
