Difference between revisions of "2017 AMC 10B Problems/Problem 22"
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===Solution 3=== | ===Solution 3=== | ||
− | As stated before, note that <math>\triangle ACB | + | As stated before, note that <math>\triangle ACB</math> is similar to <math>\triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Multiplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> <math>\frac {10\sqrt{74}}{37}</math>. We divide this by <math>2</math>, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>. |
===Solution 4=== | ===Solution 4=== |
Revision as of 20:35, 30 December 2020
Contents
Problem
The diameter of a circle of radius is extended to a point outside the circle so that . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ?
Solutions
Solution 1
Notice that and are right triangles. Then . , so . We also find that , and thus the area of is .
Solution 2
We note that by similarity. Also, since the area of and , , so the area of .
Solution 3
As stated before, note that is similar to . By similarity, we note that is equivalent to . We set to and to . By the Pythagorean Theorem, . Combining, . We can add and divide to get . We square root and rearrange to get . We know that the legs of the triangle are and . Multiplying by and eventually gives us . We divide this by , since is the formula for a triangle. This gives us .
Solution 4
Let's call the center of the circle that segment is the diameter of, . Note that is an isosceles right triangle. Solving for side , using the Pythagorean theorem, we find it to be . Calling the point where segment intersects circle , the point , segment would be . Also, noting that is a right triangle, we solve for side , using the Pythagorean Theorem, and get . Using Power of Point on point , we can solve for . We can subtract from to find and then solve for using Pythagorean theorem once more.
= (Diameter of circle + ) = =
= - =
Now to solve for :
- = + = =
Note that is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases and , we get the area of triangle to be .
Coordinate Geo
Drawing the picture, we realize that the equation for the line from A to E is , and the equation for the circle is plugging in for y we get so , that means
the height is and the base is , so the area is
-harsha12345
See Also
2017 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.