Difference between revisions of "2003 AMC 12A Problems/Problem 14"
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Since a square is a rhombus, the area of the square is <math>\frac{d_1d_2}{2}</math>, where <math>d_1</math> and <math>d_2</math> are the diagonals of the rhombus. Since the diagonal is <math>4+4\sqrt{3}</math>, the area is <math>\frac{(4+4\sqrt{3})^2}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>. | Since a square is a rhombus, the area of the square is <math>\frac{d_1d_2}{2}</math>, where <math>d_1</math> and <math>d_2</math> are the diagonals of the rhombus. Since the diagonal is <math>4+4\sqrt{3}</math>, the area is <math>\frac{(4+4\sqrt{3})^2}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>. | ||
| − | ==Solution 3== | + | ===Solution 3=== |
Because <math>ABCD</math> has area <math>16</math>, its side length is simply <math>\sqrt{16}\implies 4</math>. | Because <math>ABCD</math> has area <math>16</math>, its side length is simply <math>\sqrt{16}\implies 4</math>. | ||
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We also know that <math>KB=4, BL=4</math>. | We also know that <math>KB=4, BL=4</math>. | ||
| − | Using Law of Cosines, we find that side <math>(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sqrt{ | + | Using Law of Cosines, we find that side <math>(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sqrt{3}</math>. |
| − | However, the area of <math>KLMN</math> is simply <math>(KL)^2</math>, hence the answer is <math>32+16\sqrt{ | + | However, the area of <math>KLMN</math> is simply <math>(KL)^2</math>, hence the answer is <math>32+16\sqrt{3}\implies {D}</math>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2003|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2003|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:54, 15 June 2017
Problem
Points 
and 
lie in the plane of the square 
such that 
, 
, 
, and 
are equilateral triangles. If has an area of 16, find the area of 
.
![[asy] unitsize(2cm); defaultpen(fontsize(8)+linewidth(0.8)); pair A=(-0.5,0.5), B=(0.5,0.5), C=(0.5,-0.5), D=(-0.5,-0.5); pair K=(0,1.366), L=(1.366,0), M=(0,-1.366), N=(-1.366,0); draw(A--N--K--A--B--K--L--B--C--L--M--C--D--M--N--D--A); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$K$",K,NNW); label("$L$",L,E); label("$M$",M,S); label("$N$",N,W); [/asy]](http://latex.artofproblemsolving.com/1/4/0/14057a8f4f08bad17471fdaed56a758f1e2fcea7.png)
Solution
Solution 1
Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is 
.
The diagonal of the square KNML will then be 
. From here there are 2 ways to proceed:
First: Since the diagonal is , the side length is 
, and the area is thus 
.
Solution 2
Since a square is a rhombus, the area of the square is 
, where 
and 
are the diagonals of the rhombus. Since the diagonal is , the area is 
.
Solution 3
Because has area 
, its side length is simply 
.
Angle chasing, we find that the angle of 
.
We also know that 
.
Using Law of Cosines, we find that side 
.
However, the area of is simply 
, hence the answer is 
.
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
