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Difference between revisions of "2019 AMC 10A Problems/Problem 7"
m (Fixed formatting)
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==Solution 2==
 
==Solution 2==
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Like in Solution 1, let's first calculate the slope-intercept form of all three lines:
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Like in Solution 1, let's first work out the slope-intercept form of all three lines:
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<math>(x,y)=(2,2)</math> and <math>y=x/2 + b</math> implies <math>2=2/2 +b=1+b</math> so <math>b=1</math>, while <math>y=2x + c</math> implies <math>2= 2*2+c=4+c</math> so <math>c=-2</math>. Also, <math>x+y=10</math> implies <math>y=-x+10</math>. Thus the lines are <math>y=x/2 +1, y=2x-2,</math> and <math>y=-x+10</math>.  
+
<math>(x,y)=(2,2)</math> and <math>y=\frac{x}{2} + b</math> implies <math>2=\frac{2}{2} +b=1+b</math> so <math>b=1</math>, while <math>y=2x + c</math> implies <math>2= 2 \cdot 2+c=4+c</math> so <math>c=-2</math>. Also, <math>x+y=10</math> implies <math>y=-x+10</math>. Thus the lines are <math>y=\frac{x}{2} +1, y=2x-2,</math> and <math>y=-x+10</math>.  
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Now we find the intersections between each of the lines with <math>y=-x+10</math>, which are <math>(6,4)</math> and <math>(4,6)</math>. Applying the Shoelace Theorem, we can find that the solution is <math>6 \implies \boxed{\textbf{(C) }6}.</math>
+
Now we find the intersection points between each of the lines with <math>y=-x+10</math>, which are <math>(6,4)</math> and <math>(4,6)</math>. Applying the [[Shoelace Theorem]], we can find that the solution is <math>6 \implies \boxed{\textbf{(C) }6}.</math>
  
 
==Solution 3==
 
==Solution 3==
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==Solution 4==
 
==Solution 4==
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We can graph the points or use the methods above to find the coordinates, which are <math>(4, 6)</math>, <math>(6, 4)</math>, and <math>(2,2)</math>. Using the [[Shoelace Theorem]], we find that it is equal to <math>(4\cdot 2)-(2\cdot 6)+(2\cdot 4)-(6\cdot 2)+(6 \cdot 6)-(4\cdot 4) = 12</math>. Because the Shoelace Theorem tells us to find the half of that sum, we get <math>\boxed{\text{(C) 6}}</math>
+
We can graph the points or use the methods above to find the coordinates, which are <math>(4, 6)</math>, <math>(6, 4)</math>, and <math>(2,2)</math>. Using the [[Shoelace Theorem]], we find that it is equal to <math>(4\cdot 2)-(2\cdot 6)+(2\cdot 4)-(6\cdot 2)+(6 \cdot 6)-(4\cdot 4) = 12</math>. Because the Shoelace Theorem tells us to find the half of that sum, we get <math>\boxed{\textbf{(C) }6}</math>.
−
 
 
−
--Claire (clara32356)
 
  
 
==Solution 5 (Draw it out)==
 
==Solution 5 (Draw it out)==
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We draw the 3 lines as in the question on grid paper. It can be easily seen that the 3 points of intersection is <math>(2, 2)</math>, <math>(4, 6)</math> and <math>(6, 4)</math>. After drawing a square with vertices <math>(2, 2)</math>, <math>(2, 6)</math>, <math>(6, 6)</math> and <math>(6, 2)</math>, we get that the triangle's area is equal to <math>16-4-2-4=6</math>. Thus the answer is <math>\boxed{\text{(C) 6}}</math>
+
We draw the three lines given in the question on grid paper. It can be easily seen that the three points of intersection are <math>(2, 2)</math>, <math>(4, 6)</math> and <math>(6, 4)</math>. After drawing a square with vertices <math>(2, 2)</math>, <math>(2, 6)</math>, <math>(6, 6)</math> and <math>(6, 2)</math>, we get that the triangle's area is equal to <math>16-4-2-4=6</math>. Thus the answer is <math>\boxed{\textbf{(C) }6}</math>.
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−
-SmileKat32
 
  
 
==See Also==
 
==See Also==

Revision as of 21:06, 17 February 2019

The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.

Problem

Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$. What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$

$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution 1

The two lines are $y=2x-2$ and $y = \frac{x}{2}+1$, which intersect the third line at $(4,6)$ and $(6,4)$. So we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2} \implies \boxed{\textbf{(C) }6}$.

Solution 2

Like in Solution 1, let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$, while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$. Also, $x+y=10$ implies $y=-x+10$. Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$. Now we find the intersection points between each of the lines with $y=-x+10$, which are $(6,4)$ and $(4,6)$. Applying the Shoelace Theorem, we can find that the solution is $6 \implies \boxed{\textbf{(C) }6}.$

Solution 3

Like the other solutions, solve the systems to see that the triangles two other points are at $(4, 6)$ and $(6, 4)$. The apply Heron's Formula. The semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$. The area then reduces nicely to a difference of squares, making it $6 \implies \boxed{\textbf{(C) }6}.$

Solution 4

We can graph the points or use the methods above to find the coordinates, which are $(4, 6)$, $(6, 4)$, and $(2,2)$. Using the Shoelace Theorem, we find that it is equal to $(4\cdot 2)-(2\cdot 6)+(2\cdot 4)-(6\cdot 2)+(6 \cdot 6)-(4\cdot 4) = 12$. Because the Shoelace Theorem tells us to find the half of that sum, we get $\boxed{\textbf{(C) }6}$.

Solution 5 (Draw it out)

We draw the three lines given in the question on grid paper. It can be easily seen that the three points of intersection are $(2, 2)$, $(4, 6)$ and $(6, 4)$. After drawing a square with vertices $(2, 2)$, $(2, 6)$, $(6, 6)$ and $(6, 2)$, we get that the triangle's area is equal to $16-4-2-4=6$. Thus the answer is $\boxed{\textbf{(C) }6}$.

See Also

2019 AMC 10A (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 AMC 10 Problems and Solutions
2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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. AMC logo.png

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