Difference between revisions of "2021 AMC 12B Problems/Problem 15"

Revision as of 08:02, 12 February 2021

Problem

The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written is $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$ $[asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]$

$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$

Solution

Let $M$ be the midpoint of $CD$ . Noting that $AED$ and $ABC$ are $120-30-30$ triangles because of the equilateral triangles, $AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}$ . Also, $[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}$ and so $[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{(\textbf{D})23}$ .

~Lcz

Solution 2

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\sqrt{3}$ and $CD=2$ . Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$ , so the area is $\sqrt{11}$ . Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{(\textbf{D})23}$ .

Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)

https://youtu.be/QtSbAKUb1VE

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 32: / Line 32: @@
 ==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>\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>.
+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) ==

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