Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 10A Problems/Problem 20"

(→‎Solution 2)
 
(15 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
An equiangular octagon has four sides of length 1 and four sides of length <math>\frac{\sqrt{2}}{2}</math>, arranged so that no two consecutive sides have the same length.  What is the area of the octagon?
+
An equilangular octagon has four sides of length <math>1</math> and four sides of length <math>\frac{\sqrt{2}}{2}</math>, arranged so that no two consecutive sides have the same length.  What is the area of the octagon?
  
−
<math> \mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ }  \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ }  \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ }  \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ }  7 </math>
+
<math> \textbf{(A) } \frac72\qquad \textbf{(B) }  \frac{7\sqrt2}{2}\qquad \textbf{(C) }  \frac{5+4\sqrt2}{2}\qquad \textbf{(D) }  \frac{4+5\sqrt2}{2}\qquad \textbf{(E) }  7 </math>
  
 
==Solution 1==
 
==Solution 1==
−
The area of the octagon can be divided up into 5 squares with side <math>\frac{\sqrt2}2</math> and 4 right triangles, which are half the area of each of the squares.
+
The area of the octagon can be divided up into <math>5</math> squares with side <math>\frac{\sqrt2}2</math> and <math>4</math> right triangles, which are half the area of each of the squares.
  
 
Therefore, the area of the octagon is equal to the area of <math>5+4\left(\frac12\right)=7</math> squares.
 
Therefore, the area of the octagon is equal to the area of <math>5+4\left(\frac12\right)=7</math> squares.
  
−
The area of each square is <math>\left(\frac{\sqrt2}2\right)^2=\frac12</math>, so the area of 7 squares is <math>\frac72\Rightarrow\mathrm{\boxed{A}}</math>.
+
The area of each square is <math>\left(\frac{\sqrt2}2\right)^2=\frac12</math>, so the area of <math>7</math> squares is <math>\boxed{\textbf{(A) }\frac72}</math>.
  
 
<asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H);  draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy>
 
<asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H);  draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy>
  
 
==Solution 2==
 
==Solution 2==
−
Using the diagram from above, we can extend the sides of length <math>1</math> to form four right triangles and the octagon, all inside a square. The right triangles are 45-45-90 triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>, so the side length is <math>\frac{1}{2}</math>. Thus, the area of the larger square is <math>2 \cdot 2 = 4</math>, and the area of the four right triangles combined is <math>\frac{1}{2}</math>, so the area of the octagon is <math>\frac{7}{2}</math>, or <math> \boxed{\mathrm{(A). \ }}</math>
+
Using the diagram from above, we can extend the sides of length <math>1</math> to form four right triangles and the octagon, all inside a square. If you wanted, you could also extend the sides of length <math>\frac{\sqrt{2}}{2}</math> and get the same answer, but that would make things slightly harder.  The right triangles are <math>45-45-90</math> triangles with hypotenuse <math>\frac{\sqrt{2}}{2}</math>, so the side length is <math>\frac{1}{2}</math>. Thus, the area of the larger square is <math>2 \cdot 2 = 4</math>, and the area of the four right triangles combined is <math>\frac{1}{2}</math>, so the area of the octagon is <math>4-\frac{1}{2}=\boxed{\textbf{(A) }\frac{7}{2}}</math>
  
 +
edited by mobius247
  
 +
==Solution 3==
 +
Refer to the following diagram:
  
−
<math>4</math> - <math>\frac{1}{2}</math>
+
[[File:AMC10_2005A_P20.png|500px]]
−
<math>5+4\left(\frac12\right)=7</math>
+
 
 +
(Picture made on Geogebra)
 +
 
 +
 
 +
Note that each square has area <math>\frac14</math>, and each triangle has area <math>\frac18</math>. The total area is <math>12\cdot\frac14+4\cdot\frac18=\frac72 \Longrightarrow \boxed{\textbf{(A) } \frac72}</math>.
 +
 
 +
 
 +
Remark: This solution requires careful drawing, and also realising that connecting lines leads to squares and isosceles right triangles (notice the <math>\sqrt{2}</math> and realise that it is the hypotenuse of a <math>45-45-90</math> triangle with side length ratios <math>1:1:\sqrt{2}</math>.).
 +
 
 +
 
 +
~JH. L
  
 
==Video Solution==
 
==Video Solution==
Line 29: Line 42:
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
−
[[Category:Area Ratio Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:26, 13 October 2024

Problem

An equilangular octagon has four sides of length $1$ and four sides of length $\frac{\sqrt{2}}{2}$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?

$\textbf{(A) } \frac72\qquad \textbf{(B) } \frac{7\sqrt2}{2}\qquad \textbf{(C) } \frac{5+4\sqrt2}{2}\qquad \textbf{(D) } \frac{4+5\sqrt2}{2}\qquad \textbf{(E) } 7$

Solution 1

The area of the octagon can be divided up into $5$ squares with side $\frac{\sqrt2}2$ and $4$ right triangles, which are half the area of each of the squares.

Therefore, the area of the octagon is equal to the area of $5+4\left(\frac12\right)=7$ squares.

The area of each square is $\left(\frac{\sqrt2}2\right)^2=\frac12$, so the area of $7$ squares is $\boxed{\textbf{(A) }\frac72}$.

[asy] pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); [/asy]

Solution 2

Using the diagram from above, we can extend the sides of length $1$ to form four right triangles and the octagon, all inside a square. If you wanted, you could also extend the sides of length $\frac{\sqrt{2}}{2}$ and get the same answer, but that would make things slightly harder. The right triangles are $45-45-90$ triangles with hypotenuse $\frac{\sqrt{2}}{2}$, so the side length is $\frac{1}{2}$. Thus, the area of the larger square is $2 \cdot 2 = 4$, and the area of the four right triangles combined is $\frac{1}{2}$, so the area of the octagon is $4-\frac{1}{2}=\boxed{\textbf{(A) }\frac{7}{2}}$

edited by mobius247

Solution 3

Refer to the following diagram:

AMC10 2005A P20.png

(Picture made on Geogebra)


Note that each square has area $\frac14$, and each triangle has area $\frac18$. The total area is $12\cdot\frac14+4\cdot\frac18=\frac72 \Longrightarrow \boxed{\textbf{(A) } \frac72}$.


Remark: This solution requires careful drawing, and also realising that connecting lines leads to squares and isosceles right triangles (notice the $\sqrt{2}$ and realise that it is the hypotenuse of a $45-45-90$ triangle with side length ratios $1:1:\sqrt{2}$.).


~JH. L

Video Solution

CHECK OUT Video Solution: https://youtu.be/rwPFZnYk9V8

See Also

2005 AMC 10A (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_20&oldid=229732"