2019 AMC 10A Problems/Problem 7
- 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
The two lines are and
, which intersect the third line at
and
. So we have an isosceles triangle with base
and height
.
Solution 2
Like in Solution 1, let's first calculate 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 intersections between each of the lines with
, which are
and
. Applying the Shoelace Theorem, we can find that the solution is
Solution 3
Like the other solutions, solve the systems to see that the triangles two other points are at and
. The apply Heron's Formula. The semi-perimeter will be
. The area then reduces nicely to a difference of squares, making it
Solution 4
We can graph the points or use the methods above to find the coordinates, which are ,
, and
. Using the Shoelace Theorem, we find that it is equal to
. Because the Shoelace Theorem tells us to find the half of that sum, we get
--Claire (clara32356)
Solution 5 (Draw it out)
We draw the 3 lines as in the question on grid paper. It can be easily seen that the 3 points of intersection is ,
and
. After drawing a square with vertices
,
,
and
, we get that the triangle's area is equal to
. Thus the answer is
-SmileKat32
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 |
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