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 "2015 AMC 8 Problems/Problem 21"

m (→‎Problem)
(→‎Solution 2)
 
(2 intermediate revisions by 2 users not shown)
Line 26: Line 26:
 
<math>\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32</math>.
 
<math>\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32</math>.
  
−
== Solution ==
+
== Solution 1 ==
  
 
Clearly, since <math>\overline{FE}</math> is a side of a square with area <math>32</math>, <math>\overline{FE} = \sqrt{32} = 4 \sqrt{2}</math>. Now, since <math>\overline{FE} = \overline{BC}</math>, we have <math>\overline{BC} = 4 \sqrt{2}</math>.
 
Clearly, since <math>\overline{FE}</math> is a side of a square with area <math>32</math>, <math>\overline{FE} = \sqrt{32} = 4 \sqrt{2}</math>. Now, since <math>\overline{FE} = \overline{BC}</math>, we have <math>\overline{BC} = 4 \sqrt{2}</math>.
Line 33: Line 33:
  
 
Lastly, <math>\Delta KBC</math> is a right triangle. We see that <math>\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}</math>, so <math>\Delta KBC</math> is a right triangle with legs <math>3 \sqrt{2}</math> and <math>4 \sqrt{2}</math>. Now, its area is <math>\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{\textbf{(C)}~12}</math>.
 
Lastly, <math>\Delta KBC</math> is a right triangle. We see that <math>\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}</math>, so <math>\Delta KBC</math> is a right triangle with legs <math>3 \sqrt{2}</math> and <math>4 \sqrt{2}</math>. Now, its area is <math>\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{\textbf{(C)}~12}</math>.
 +
 +
== Solution 2 ==
 +
Since <math>\overline{FE} = \sqrt{32}</math>, and <math>\overline{FE} = \overline{BC}</math>, <math>\overline{BC} = 4\sqrt{2}</math>. Meanwhile, <math>\overline{JB} = 3\sqrt{2}</math>, and since <math>\triangle JBK</math> is equilateral, <math>\overline{BK} = 3\sqrt{2}</math>. If <math>ABCDEF</math> is equiangular, <math>\angle ABC = \frac{180 \cdot (n-2)}{n} = 120^{\circ}</math>, where <math>n</math> is the number of sides of the shape. Adding all the angles around <math>B</math> gives <math>270^{\circ}</math>, so <math>\angle KBC = 360 - 90 = 270^{\circ}</math>. Because <math>\triangle KBC</math> is right, the area of <math>\triangle KBC = \frac{4\sqrt{2} \cdot 3\sqrt{2}} {2} = 12</math>. Therefore, the answer is <math>\boxed{\textbf{(C)}~12}</math>. ~strongstephen
 +
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/mUBhWTlvuLQ
 
https://youtu.be/mUBhWTlvuLQ

Latest revision as of 19:36, 17 January 2024

Problem

In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$?

[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]

$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$.

Solution 1

Clearly, since $\overline{FE}$ is a side of a square with area $32$, $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$. Now, since $\overline{FE} = \overline{BC}$, we have $\overline{BC} = 4 \sqrt{2}$.

Now, $\overline{JB}$ is a side of a square with area $18$, so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$. Since $\Delta JBK$ is equilateral, $\overline{BK} = 3 \sqrt{2}$.

Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$, so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$. Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{\textbf{(C)}~12}$.

Solution 2

Since $\overline{FE} = \sqrt{32}$, and $\overline{FE} = \overline{BC}$, $\overline{BC} = 4\sqrt{2}$. Meanwhile, $\overline{JB} = 3\sqrt{2}$, and since $\triangle JBK$ is equilateral, $\overline{BK} = 3\sqrt{2}$. If $ABCDEF$ is equiangular, $\angle ABC = \frac{180 \cdot (n-2)}{n} = 120^{\circ}$, where $n$ is the number of sides of the shape. Adding all the angles around $B$ gives $270^{\circ}$, so $\angle KBC = 360 - 90 = 270^{\circ}$. Because $\triangle KBC$ is right, the area of $\triangle KBC = \frac{4\sqrt{2} \cdot 3\sqrt{2}} {2} = 12$. Therefore, the answer is $\boxed{\textbf{(C)}~12}$. ~strongstephen

Video Solution

https://youtu.be/mUBhWTlvuLQ

~savannahsolver

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 AJHSME/AMC 8 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=2015_AMC_8_Problems/Problem_21&oldid=211067"