Difference between revisions of "2017 AMC 8 Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | + | In the figure shown, <math>\overline{US}</math> and <math>\overline{UT}</math> are line segments each of length 2, and <math>m\angle TUS = 60^\circ</math>. Arcs <math>\overarc{TR}</math> and <math>\overarc{SR}</math> are each one-sixth of a circle with radius 2. What is the area of the region shown? | |
<asy>draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);</asy> | <asy>draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);</asy> |
Revision as of 00:28, 2 December 2023
Contents
[hide]Problem
In the figure shown, and
are line segments each of length 2, and
. Arcs
and
are each one-sixth of a circle with radius 2. What is the area of the region shown?
Solution 1
Let the centers of the circles containing arcs and
be
and
, respectively. Extend
and
to
and
, and connect point
with point
.
We can clearly see that
is an equilateral triangle, because the problem states that
. We can figure out that
and
because they are
of a circle. The area of the figure is equal to
minus the combined area of the
sectors of the circles (in red). Using the area formula for an equilateral triangle,
where
is the side length of the equilateral triangle,
is
The combined area of the
sectors is
, which is
Thus, our final answer is
Solution 2
In addition to the given diagram, we can draw lines and
The area of rhombus
is half the product of its diagonals, which is
. However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by
, then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is
The area of rhombus
minus the circular segments is
~PEKKA
Video Solutions
- Happytwin
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=1350
~ pi_is_3.14
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.