Difference between revisions of "2005 AMC 8 Problems/Problem 23"
Littlelachy (talk | contribs) m (→Easier and More Logical Solution) |
m (/* Solution 2 *\) |
||
| Line 19: | Line 19: | ||
==Easier and More Logical Solution== | ==Easier and More Logical Solution== | ||
| − | + | First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be 4*4 = 16. Divide 16 by 2 to get he original shape and you get <math>\boxed{8}</math> | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=22|num-a=24}} | {{AMC8 box|year=2005|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:22, 10 July 2017
Problem
Isosceles right triangle 
encloses a semicircle of area 
. The circle has its center 
on hypotenuse 
and is tangent to sides 
and 
. What is the area of triangle ?
![[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]](http://latex.artofproblemsolving.com/8/a/7/8a7fba91b0cdfee41e69cbd89d6b1389d4169877.png)
Solution
The semi circle has an area of 
and a radius of 
.
Because this is an isosceles right triangle, the center is the midpoint of the hypotenuse. Radii drawn to the tangent points of the semicircle and the radii also divide the legs into two equal segments. They also create a square in the top left corner. From this, we can conclude the legs of the triangle are twice the length of the radii, 
. The area of the triangle is 
.
Easier and More Logical Solution
First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be 4*4 = 16. Divide 16 by 2 to get he original shape and you get 
See Also
| 2005 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
