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Difference between revisions of "1992 AHSME Problems/Problem 9"

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== Solution ==
 
== Solution ==
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>
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First, we calculate the area of 1 triangle. For an equilateral triangle with side s, its area is <math>\frac\{\sqrt{3}s^{2}}{4}</math>. If the side of the equilateral triangle is <math>\2\sqrt{3}</math>, the area of one such triangle is <math>\3\sqrt{3}</math>.
+
First, we calculate the area of 1 triangle. For an equilateral triangle with side <math>s</math>, its area is <math>\frac{\sqrt{3}s^{2}}{4}</math>. If the side of the equilateral triangle is <math>2\sqrt{3}</math>, the area of one such triangle is <math>3\sqrt{3}</math>.
 
There are 5 equilateral triangles in total, overlapping by an area of 4 smaller equilateral triangles. Each smaller triangle is one fourth as big as the big equilateral triangles. Therefore, we subtract the overlapping area, which is equivalent to the area of 1 big equilateral triangle.  
 
There are 5 equilateral triangles in total, overlapping by an area of 4 smaller equilateral triangles. Each smaller triangle is one fourth as big as the big equilateral triangles. Therefore, we subtract the overlapping area, which is equivalent to the area of 1 big equilateral triangle.  
−
Hence, the total area is equal to the area of 4 equilateral triangles, which is <math>\3\sqrt{3} * 4 = 12\sqrt{3} </math>.
+
Hence, the total area is equal to the area of 4 equilateral triangles, which is <math>3\sqrt{3} \times 4 = 12\sqrt{3} </math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 01:29, 20 February 2018

Problem

[asy] draw((-7,0)--(7,0),black+linewidth(.75)); draw((-3*sqrt(3),0)--(-2*sqrt(3),3)--(-sqrt(3),0)--(0,3)--(sqrt(3),0)--(2*sqrt(3),3)--(3*sqrt(3),0),black+linewidth(.75)); draw((-2*sqrt(3),0)--(-1*sqrt(3),3)--(0,0)--(sqrt(3),3)--(2*sqrt(3),0),black+linewidth(.75)); [/asy]

Five equilateral triangles, each with side $2\sqrt{3}$, are arranged so they are all on the same side of a line containing one side of each vertex. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is

$\text{(A) 10} \quad \text{(B) } 12\quad \text{(C) } 15\quad \text{(D) } 10\sqrt{3}\quad \text{(E) } 12\sqrt{3}$

Solution

$\fbox{E}$ First, we calculate the area of 1 triangle. For an equilateral triangle with side $s$, its area is $\frac{\sqrt{3}s^{2}}{4}$. If the side of the equilateral triangle is $2\sqrt{3}$, the area of one such triangle is $3\sqrt{3}$. There are 5 equilateral triangles in total, overlapping by an area of 4 smaller equilateral triangles. Each smaller triangle is one fourth as big as the big equilateral triangles. Therefore, we subtract the overlapping area, which is equivalent to the area of 1 big equilateral triangle. Hence, the total area is equal to the area of 4 equilateral triangles, which is $3\sqrt{3} \times 4 = 12\sqrt{3}$.

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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