Difference between revisions of "2014 AMC 10B Problems/Problem 13"
(→Problem: Added question w/o diagram) |
Cdl031apos (talk | contribs) m (→Solution 2) |
||
(15 intermediate revisions by 6 users not shown) | |||
Line 3: | Line 3: | ||
Six regular hexagons surround a regular hexagon of side length <math>1</math> as shown. What is the area of <math>\triangle{ABC}</math>? | Six regular hexagons surround a regular hexagon of side length <math>1</math> as shown. What is the area of <math>\triangle{ABC}</math>? | ||
− | ( | + | <asy> |
+ | draw((0,0)--(-5,8.66025404)--(0, 17.3205081)--(10, 17.3205081)--(15,8.66025404)--(10, 0)--(0, 0)); | ||
+ | draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0)); | ||
+ | draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0)); | ||
+ | draw((0,0)--(-5, -8.66025404)--(0, -17.3205081)--(10, -17.3205081)--(15, -8.66025404)--(10, 0)--(0, 0)); | ||
+ | draw((15,8.66025404)--(10, 17.3205081)--(15, 25.9807621)--(25, 25.9807621)--(30, 17.3205081)--(25, 8.66025404)--(15, 8.66025404)); | ||
+ | draw((15,-8.66025404)--(10, -17.3205081)--(15, -25.9807621)--(25, -25.9807621)--(30, -17.3205081)--(25, -8.66025404)--(15, -8.66025404)); | ||
+ | label("A", (0,0), W); | ||
+ | label("B", (30, 17.3205081), NE); | ||
+ | label("C", (30, -17.3205081), SE); | ||
+ | draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | //(Diagram Creds-DivideBy0) | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | |||
+ | <math> \textbf {(A) } 2\sqrt{3} \qquad \textbf {(B) } 3\sqrt{3} \qquad \textbf {(C) } 1+3\sqrt{2} \qquad \textbf {(D) } 2+2\sqrt{3} \qquad \textbf {(E) } 3+2\sqrt{3} </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
+ | |||
+ | We note that the <math>6</math> triangular sections in <math>\triangle{ABC}</math> can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as <math>\dfrac{3\sqrt{3}}{2}</math>. The area of <math>\triangle{ABC}</math>, which is equivalent to two of these hexagons together, is <math>\boxed{\textbf{(B)} 3\sqrt{3}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | The measure of an interior angle in a hexagon is 120 degrees. Each side of the 6 triangles that make up the remainder of triangle ABC are isosceles with 2 side lengths of 1 and an angle of 120 degrees. Therefore, by the Law of Cosines, we calculate that the longest side of this triangle is <math>\sqrt{3}</math>, so the side length of triangle ABC is <math>2\sqrt{3}</math>. Using the equilateral triangle area formula, we figure out that the answer is <math>\boxed{\textbf{(B)} 3\sqrt{3}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2014|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:35, 11 July 2020
Contents
Problem
Six regular hexagons surround a regular hexagon of side length as shown. What is the area of ?
Solution
We note that the triangular sections in can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as . The area of , which is equivalent to two of these hexagons together, is .
Solution 2
The measure of an interior angle in a hexagon is 120 degrees. Each side of the 6 triangles that make up the remainder of triangle ABC are isosceles with 2 side lengths of 1 and an angle of 120 degrees. Therefore, by the Law of Cosines, we calculate that the longest side of this triangle is , so the side length of triangle ABC is . Using the equilateral triangle area formula, we figure out that the answer is .
See Also
2014 AMC 10B (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.