Difference between revisions of "2006 AMC 10B Problems/Problem 19"
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== Solutions == | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
− | The shaded area is equivalent to the area of sector <math>DOE</math> | + | The shaded area is equivalent to the area of sector <math>DOE</math> minus the area of triangle <math>DOE</math> plus the area of triangle <math>DBE</math>. |
Using the [[Pythagorean Theorem]], <math>(DA)^2=(CE)^2=2^2-1^2=3</math> so <math>DA=CE=\sqrt{3}</math>. | Using the [[Pythagorean Theorem]], <math>(DA)^2=(CE)^2=2^2-1^2=3</math> so <math>DA=CE=\sqrt{3}</math>. |
Revision as of 17:05, 2 June 2021
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
?
Solutions
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
Non-Trig Approach
has the same height as
which is
We already know that
Therefore the area is
Since
Therefore the sum of the areas is
Then the area of the shaded area becomes
~mathboy282
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
Solution 3 (Using Answer Choices)
Like the first solutions, you find that the area of sector is
. We also know that the triangles will not be in terms of
. Looking at the answers, choices
and
both contain
. However, based on the diagram, we observe that the answer must be less than
. Only
consists of a value less than
.
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.