2014 AMC 10A Problems/Problem 13
Contents
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,
Solution 4
We know that the area is equal to 3*EAF+3*ACGF+ABC. We also know that ACGF and the rest of the squares' area is equal to 1. Therefore the answer is 3*EAF+ABC+3. The only one with "+3" or "3+" is C, our answer. Very unreliable. -Reality Writes
Solution 5
The area of the obtuse triangle is
The total area is
~mathboy282
Solution 6
The total area is the sum of the three squares, the three (congruent) obtuse triangles, and the equilateral triangle. The area of the equilateral triangle is and the area of each square is . The area of a triangle in general is where and are two sides and is the included angle. measures because and are right, and . So the area of the obtuse triangle is . The total area is .
~JH. L
Solution 7
Since , Applying the Law of Cosines on gives us Since is isosceles, the perpendicular bisector of also intersects segment in its median, which we can call point Hence, we can apply the Pythagorean theorem on or to get We can use this to get the area of the triangle and multiply that by three since the triangles are congruent. The result follows. ~peelybonehead
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AMC 10 Problems and Solutions |
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