2020 AMC 10A Problems/Problem 12
- 1 Problem 12
- 2 Solution 1
- 3 Solution 2 (Trapezoid)
- 4 Solution 3 (Medians)
- 5 Solution 4 (Triangles)
- 6 Solution 5 (Only Pythagorean Theorem)
- 7 Solution 6 (Drawing)
- 8 Solution 7
- 9 Solution 8 (Fastest)
- 10 Solution 9
- 11 Solution 10 (High IQ)
- 12 Solution 11 (Kite)
- 13 Solution 12 (Educated guessing)
- 14 Video Solution 1
- 15 Video Solution 2
- 16 Video Solution 3
- 17 Video Solution 4 by OmegaLearn
- 18 See Also
Triangle is isosceles with . Medians and are perpendicular to each other, and . What is the area of
Since quadrilateral has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that has the area of triangle by similarity, so Thus,
Solution 2 (Trapezoid)
We know that , and since the ratios of its sides are , the ratio of of their areas is .
If is the area of , then trapezoid is the area of .
Let's call the intersection of and . Let . Then . Since , and are heights of triangles and , respectively. Both of these triangles have base .
Adding these two gives us the area of trapezoid , which is .
This is of the triangle, so the area of the triangle is ~quacker88, diagram by programjames1
Solution 3 (Medians)
Draw median .
Since we know that all medians of a triangle intersect at the centroid, we know that passes through point . We also know that medians of a triangle divide each other into segments of ratio . Knowing this, we can see that , and since the two segments sum to , and are and , respectively.
Finally knowing that the medians divide the triangle into sections of equal area, finding the area of is enough. .
The area of . Multiplying this by gives us
Solution 4 (Triangles)
We know that , , so .
As , we can see that and with a side ratio of .
So , .
With that, we can see that , and the area of trapezoid is 72.
As said in solution 1, .
-QuadraticFunctions, solution 1 by ???
Solution 5 (Only Pythagorean Theorem)
Let be the height. Since medians divide each other into a ratio, and the medians have length 12, we have and . From right triangle , so . Since is a median, . From right triangle , which implies . By symmetry .
Applying the Pythagorean Theorem to right triangle gives , so . Then the area of is
Solution 6 (Drawing)
By similarity, the area of is equal to .
The area of is equal to 72.
Assuming the total area of the triangle is S, the equation will be : S = 72.
Given a triangle with perpendicular medians with lengths and , the area will be .
Solution 8 (Fastest)
Connect the line segment and it's easy to see quadrilateral has an area of the product of its diagonals divided by which is . Now, solving for triangle could be an option, but the drawing shows the area of will be less than the quadrilateral meaning the the area of is less than but greater than , leaving only one possible answer choice, .
Connect , and let be the point where intersects . because all medians of a triangle intersect at one point, which in this case is . because the point at which all medians intersect divides the medians into segments of ratio , so and similarly . We apply the Pythagorean Theorem to triangle and get . The area of triangle is , and that must equal to , so . , so . The area of triangle is equal to .
Solution 10 (High IQ)
. Let intersection of line and base be
Solution 11 (Kite)
Since and intersect at a right angle, this means is a kite. Hence, the area of this kite is .
Also, notice that by AAA Similarity. Since the ratios of its sides is , the ratios of the area is . Therefore,
Simplifying gives us , so
Solution 12 (Educated guessing)
Horizontally translate line until point is at point , with subsequently at , and then connect up and to create , which is a right triangle.
Notice that = , and = (since the latter has the same base and height as the sub-triangle inside ).
From this, we can deduce that cannot be true, since an incomplete part of is equal to it. We can also deduce that also cannot be true, since the unknown triangle , and . Therefore, the answer must be between and , leaving as the only possible correct answer.
Video Solution 1
Education, The Study of Everything
Video Solution 2
Video Solution 3
Video Solution 4 by OmegaLearn
|2020 AMC 10A (Problems • Answer Key • Resources)|
|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|