Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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− | <math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}</math> - <math>\frac{1}{3}</math> = <math>\frac{1}{6}</math>. <math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle ADB</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2</math> = <math>\frac{\sqrt{2}}{3}</math>. Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. Hence the height of <math>\triangle DCE</math> is <math>\frac{frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}</math> | + | <math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}</math> - <math>\frac{1}{3}</math> = <math>\frac{1}{6}</math>. <math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle ADB</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2</math> = <math>\frac{\sqrt{2}}{3}</math>. Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. Hence the height of <math>\triangle DCE</math> is <math>\frac{frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}</math> |
==See also== | ==See also== |
Revision as of 22:25, 24 December 2008
Problem
Let be a diameter of a circle and let be a point on with . Let and be points on the circle such that and is a second diameter. What is the ratio of the area of to the area of ?
. The area of
Solution
http://img443.imageshack.us/img443/8034/circlenc1.png
is of diameter and is - = . is the radius of the circle, so using the Pythagorean theorem height of is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2$ (Error compiling LaTeX. ! Missing } inserted.) = . Area of the is = . The height of can be found using the area of and as base. Hence the height of is
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |