Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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4&2&0\\ | 4&2&0\\ | ||
\end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}</cmath> | \end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}</cmath> | ||
+ | |||
+ | ==Solution 9== | ||
+ | Like in other solutions, we find that the three points of intersection are <math>(2, 2)</math>, <math>(4, 6)</math>, and <math>(6, 4)</math>. By the Pythagorean theorem, this is an isosceles triangle with base <math>2\sqrt2</math> and equal length <math>2\sqrt5</math>. The area of an isosceles triangle with base <math>b</math> and equal length <math>l</math> is <math>\frac{b\sqrt{4l^2-b^2}}{4}</math>. Plugging in <math>b = 2\sqrt2</math> and <math>l = 2\sqrt5</math>, | ||
+ | <cmath>\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{\textbf{(C) }6}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 23:18, 29 March 2019
- The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.
Contents
Problem
Two lines with slopes and
intersect at
. What is the area of the triangle enclosed by these two lines and the line
Solution 1
Let's first work out the slope-intercept form of all three lines:
and
implies
so
, while
implies
so
. Also,
implies
. Thus the lines are
and
.
Now we find the intersection points between each of the lines with
, which are
and
. Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base
and height
, whose area is
.
Solution 2
Like in Solution 1, we determine the coordinates of the three vertices of the triangle. Now, using the Shoelace Theorem, we can directly find that the area is .
Solution 3
Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at and
. Then apply Heron's Formula: the semi-perimeter will be
, so the area reduces nicely to a difference of squares, making it
.
Solution 4
Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are ,
, and
. We can now draw the bounding square with vertices
,
,
and
, and deduce that the triangle's area is
.
Solution 5
Like in other solutions, we find that the three points of intersection are ,
, and
. Using graph paper, we can see that this triangle has
boundary lattice points and
interior lattice points. By Pick's Theorem, the area is
.
Solution 6
Like in other solutions, we find the three points of intersection. Label these ,
, and
. By the Pythagorean Theorem,
and
. By the Law of Cosines,
Therefore,
, so the area is
.
Solution 7
Like in other solutions, we find that the three points of intersection are ,
, and
. The area of the triangle is half the absolute value of the determinant of the matrix determined by these points.
Solution 8
Like in other solutions, we find the three points of intersection. Label these ,
, and
. Then vectors
and
. The area of the triangle is half the magnitude of the cross product of these two vectors.
Solution 9
Like in other solutions, we find that the three points of intersection are ,
, and
. By the Pythagorean theorem, this is an isosceles triangle with base
and equal length
. The area of an isosceles triangle with base
and equal length
is
. Plugging in
and
,
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.