Difference between revisions of "1987 AHSME Problems/Problem 22"
(Created page with "==Problem== A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice), leaving a hole <math>24</math> cm across as the top and <math>8<...") |
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\textbf{(C)}\ 13 \qquad | \textbf{(C)}\ 13 \qquad | ||
\textbf{(D)}\ 8\sqrt{3} \qquad | \textbf{(D)}\ 8\sqrt{3} \qquad | ||
| − | \textbf{(E)}\ 6\sqrt{6} </math> | + | \textbf{(E)}\ 6\sqrt{6} </math> |
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| + | ==Solution== | ||
| + | Consider a cross-section of this problem in which a circle lies with it's center somewhere above a line. A line segment of <math>8</math> cm can be drawn from the line to the bottom of the ball. Denote the distance between the center of the circle and the line as <math>x</math>. We can construct a right triangle by dragging the center of the circle to the intersection of the circle and the line. We then have the equation <math>x^2+(12)^2=(x+8)^2</math>, <math>x^2+144=x^2+16x+64</math>. Solving, the answer is <math>\textbf{(C)}\ 13 \qquad</math> | ||
== See also == | == See also == | ||
Revision as of 21:30, 14 February 2018
Problem
A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice),
leaving a hole 
cm across as the top and 
cm deep. What was the radius of the ball (in centimeters)?
Solution
Consider a cross-section of this problem in which a circle lies with it's center somewhere above a line. A line segment of cm can be drawn from the line to the bottom of the ball. Denote the distance between the center of the circle and the line as 
. We can construct a right triangle by dragging the center of the circle to the intersection of the circle and the line. We then have the equation 
, 
. Solving, the answer is 
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
