Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that <math>x</math> is the length of the line from B to D. So, the perimeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find out <math>x</math> from these two equations. We can find out that <math>x = 2</math>, so that means that the area of <math>\triangle{ABD} = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math> | Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that <math>x</math> is the length of the line from B to D. So, the perimeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find out <math>x</math> from these two equations. We can find out that <math>x = 2</math>, so that means that the area of <math>\triangle{ABD} = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math> | ||
+ | THIS SOLUTION MAKES NO SENSE PLEASE MOVE ON | ||
==Solution 2== | ==Solution 2== |
Revision as of 19:27, 9 November 2018
Problem 16
In the figure below, choose point on so that and have equal perimeters. What is the area of ?
Solution 1
Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that is the length of the line from B to D. So, the perimeter of would be , while the perimeter of would be . Notice that we can find out from these two equations. We can find out that , so that means that the area of
THIS SOLUTION MAKES NO SENSE PLEASE MOVE ON
Solution 2
We know that the perimeters of the two small triangles are and . Setting both equal and using , we have and . Now, we simply have to find the area of . Since , we must have . Combining this with the fact that , we get
Solution 3
Since point is on line , it will split it into and . Let and . Triangle has side lengths and triangle has side lengths . Since both perimeters are equal, we have the equation . Eliminating and solving the resulting linear equation gives . Draw a perpendicular from point to . Call the point of intersection . Because angle is common to both triangles and , and both are right triangles, both are similar. The hypotenuse of triangle is 2, so the altitude must be Because and share the same altitude, the height of therefore must be . The base of is 4, so
Solution 4
Using any preferred method, realize . Since we are given a 3-4-5 right triangle, we know the value of . Since we are given , apply the Sine Area Formula to get .
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
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