Difference between revisions of "2006 AMC 10B Problems/Problem 19"
Firebolt360 (talk | contribs) (→Solution 2) |
Firebolt360 (talk | contribs) (→Solution 2) |
||
Line 59: | Line 59: | ||
label("$D$", D, dir(point--D)); | label("$D$", D, dir(point--D)); | ||
label("$E$", E, dir(point--E)); | label("$E$", E, dir(point--E)); | ||
− | label("$B$", | + | label("$B$", (1.33,1.04), SW);</asy> |
Revision as of 19:00, 15 September 2018
Contents
Problem
A circle of radius is centered at
. Square
has side length
. Sides
and
are extended past
to meet the circle at
and
, respectively. What is the area of the shaded region in the figure, which is bounded by
,
, and the minor arc connecting
and
?
Solution 1
The shaded area is equivalent to the area of sector , minus the area of triangle
plus the area of triangle
.
Using the Pythagorean Theorem, so
.
Clearly, and
are
triangles with
. Since
is a square,
.
can be found by doing some subtraction of angles.
So, the area of sector is
.
The area of triangle is
.
Since ,
. So, the area of triangle
is
. Therefore, the shaded area is
Solution 2
From the pythagorean theorem, we can see that is
. Then,
. The area of the shaded element is the area of sector
minus the areas of triangle
and triangle
combined. Below is an image to help.
Using the Base Altitude formula, where and
are the bases and
and
are the altitudes, respectively,
. The area of sector
is
of circle
. The area of circle
is
, and therefore we have the area of sector
to be
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.