Difference between revisions of "2014 AMC 10A Problems/Problem 13"
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We will use, <math>\frac{1}{2}ab\sin x</math> to find the area of the following triangles. Since <math>\angle A=360</math>, <math>\angle EAF=360-90-90-60=120</math>. | We will use, <math>\frac{1}{2}ab\sin x</math> to find the area of the following triangles. Since <math>\angle A=360</math>, <math>\angle EAF=360-90-90-60=120</math>. | ||
− | The Area of <math>\triangle AEF</math> is <math>\frac{1}{2} \cdot 1 \cdot 1 \cdot sin 120</math>. Noting, <math>sin (2x) = | + | The Area of <math>\triangle AEF</math> is <math>\frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(120)</math>. Noting, <math>\sin (2x) = 2\sin (x)\cos (x)</math>, |
− | Area of <math>\triangle AEF = \frac{1}{2} \cdot 1 \cdot 1 \cdot 2 \cdot sin(60) \cdot cos(60) = \dfrac{\sqrt{3}}{4}</math>, | + | Area of <math>\triangle AEF = \frac{1}{2} \cdot 1 \cdot 1 \cdot 2 \cdot \sin(60) \cdot \cos(60) = \dfrac{\sqrt{3}}{4}</math>, |
− | Area of <math>\triangle ABC = \frac{1}{2} \cdot 1 \cdot 1 \cdot sin(60) = \dfrac{\sqrt{3}}{4}</math>, | + | Area of <math>\triangle ABC = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(60) = \dfrac{\sqrt{3}}{4}</math>, |
Area of square ABDE = 1, | Area of square ABDE = 1, |
Revision as of 23:01, 19 October 2019
Problem
Equilateral has side length
, and squares
,
,
lie outside the triangle. What is the area of hexagon
?
Solution 1
The area of the equilateral triangle is . The area of the three squares is
.
Since ,
.
Dropping an altitude from to
allows to create a
triangle since
is isosceles. This means that the height of
is
and half the length of
is
. Therefore, the area of each isosceles triangle is
. Multiplying by
yields
for all three isosceles triangles.
Therefore, the total area is .
Solution 2
As seen in the previous solution, segment is
. Think of the picture as one large equilateral triangle,
with the sides of
, by extending
,
, and
to points
,
, and
, respectively. This makes the area of
.
Triangles ,
, and
have sides of
, so their total area is
.
Now, you subtract their total area from the area of :
Solution 3
We will use, to find the area of the following triangles. Since
,
.
The Area of is
. Noting,
,
Area of ,
Area of ,
Area of square ABDE = 1,
Therefore the composite area of the entire figure is,
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.