Difference between revisions of "2021 AMC 12B Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <math>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}</math>. Also, <math>[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}</math> and so <math>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{ | + | Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <math>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}</math>. Also, <math>[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}</math> and so <math>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}</math>. |
~Lcz | ~Lcz | ||
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==Solution 2== | ==Solution 2== | ||
− | Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{ | + | Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>. |
== Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) == | == Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) == |
Revision as of 08:04, 12 February 2021
Contents
[hide]Problem
The figure is constructed from line segments, each of which has length . The area of pentagon can be written is , where and are positive integers. What is
Solution
Let be the midpoint of . Noting that and are triangles because of the equilateral triangles, . Also, and so .
~Lcz
Solution 2
Draw diagonals and to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles and , they each have area . For triangle , we can see that and . Using Pythagorean Theorem, the altitude for this triangle is , so the area is . Adding each part up, we get .
Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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