Difference between revisions of "2009 AMC 8 Problems/Problem 7"
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The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD? | The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD? | ||
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<asy> | <asy> | ||
size(250); | size(250); | ||
| Line 35: | Line 36: | ||
label(rotate(63.43494882)*"Brown", A--D, NW); | label(rotate(63.43494882)*"Brown", A--D, NW); | ||
</asy> | </asy> | ||
| − | <math>\textbf{(A)}\ 2\qquad | + | |
| − | \textbf{(B)}\ 3 \qquad | + | <math>\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math> |
| − | \textbf{(C)}\ 4.5 \qquad | ||
| − | \textbf{(D)}\ 6 \qquad | ||
| − | \textbf{(E)}\ 9</math> | ||
==Solution== | ==Solution== | ||
The area of a triangle is <math>\frac12 bh</math>. If we let <math>CD</math> be the base of the triangle, then the height is <math>AB</math>, and the area is <math>\frac12 \cdot 3 \cdot 3 = \boxed{\textbf{(C)}\ 4.5}</math>. | The area of a triangle is <math>\frac12 bh</math>. If we let <math>CD</math> be the base of the triangle, then the height is <math>AB</math>, and the area is <math>\frac12 \cdot 3 \cdot 3 = \boxed{\textbf{(C)}\ 4.5}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can see that there is a big triangle encasing <math>ACD</math>. The area of that triangle is <math>\frac12 \cdot 3 \cdot 6 = 9</math>. We can easily see that triangle <math>ABC</math> is <math>\frac12 \cdot 3 \cdot 3</math>, which is <math>4.5</math>. <math>9-4.5</math> is <math>4.5</math>, so the answer is <math>\boxed{\textbf{(C)}\ 4.5}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=Opz71P5o4uI ~David | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/j3QSD5eDpzU?t=158 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/QfQIqS60m5s | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=6|num-a=8}} | {{AMC8 box|year=2009|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 07:23, 30 May 2023
Contents
Problem
The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
![[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]](http://latex.artofproblemsolving.com/5/b/3/5b3fa3817ad8d7032d6357d1446ffb5cfec46ac6.png)
Solution
The area of a triangle is 
. If we let 
be the base of the triangle, then the height is 
, and the area is 
.
Solution 2
We can see that there is a big triangle encasing 
. The area of that triangle is 
. We can easily see that triangle 
is 
, which is 
. 
is , so the answer is 
.
Video Solution
https://www.youtube.com/watch?v=Opz71P5o4uI ~David
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=158
~ pi_is_3.14
Video Solution by WhyMath
~savannahsolver
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
