Difference between revisions of "2021 AMC 12A Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | Trapezoid <math>ABCD</math> has <math>\overline{AB}\parallel\overline{CD},BC=CD=43</math>, and <math>\overline{AD}\perp\overline{BD}</math>. Let <math>O</math> be the intersection of the diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>, and let <math>P</math> be the midpoint of <math>\overline{BD}</math>. Given that <math>OP=11</math>, the length of <math>AD</math> can be written in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the | + | Trapezoid <math>ABCD</math> has <math>\overline{AB}\parallel\overline{CD},BC=CD=43</math>, and <math>\overline{AD}\perp\overline{BD}</math>. Let <math>O</math> be the intersection of the diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math>, and let <math>P</math> be the midpoint of <math>\overline{BD}</math>. Given that <math>OP=11</math>, the length of <math>AD</math> can be written in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. What is <math>m+n</math>? |
<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> |
Revision as of 01:34, 12 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 square 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 |
2021 AMC 10A (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 10 Problems and Solutions |
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