Difference between revisions of "2020 AMC 10A Problems/Problem 20"
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==Solution 4 (Trignometry)== | ==Solution 4 (Trignometry)== | ||
− | Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, we find <math>BE=3\ | + | Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>, so <math>[ABCD]=[ACD]+\frac{1/5}{[ACD]}=300+60=\boxed {\textbf{D) }360}</math> |
(This solution is incomplete, can someone complete it please) | (This solution is incomplete, can someone complete it please) | ||
+ | Edited by kc5170 | ||
==Video Solution== | ==Video Solution== |
Revision as of 14:20, 1 February 2020
Contents
Problem
Quadrilateral satisfies
and
Diagonals
and
intersect at point
and
What is the area of quadrilateral
Solution 1
It's crucial to draw a good diagram for this one. Since and
, we get
. Now we need to find
to get the area of the whole quadrilateral. Drop an altitude from
to
and call the point of intersection
. Let
. Since
, then
. By dropping this altitude, we can also see two similar triangles,
and
. Since
is
, and
, we get that
. Now, if we redraw another diagram just of
, we get that
. Now expanding, simplifying, and dividing by the GCF, we get
. This factors to
. Since lengths cannot be negative,
. Since
,
. So
(I'm very sorry if you're a visual learner)
~Ultraman
Solution 2 (Pro Guessing Strats)
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to and
, and because the hypotenuse is 20 we get
. Testing small numbers, we get that when
and
,
is indeed a square. The area of the triangle is thus 60, so the answer is
.
~tigershark22 ~(edited by HappyHuman)
Solution 3 (coordinates)
Let the points be
,
,
,
,and
, respectively. Since
lies on line
, we know that
. Furthermore, since
,
lies on the circle with diameter
, so
. Solving for
and
with these equations, we get the solutions
and
. We immediately discard the
solution as
should be negative. Thus, we conclude that
.
Solution 4 (Trignometry)
Using the law of cosines, express and
in terms of
. The sum of these two equations is
by the Pythagorean Theorem. Solving for
, we find
. Since
and
,
, which is five times
, so
(This solution is incomplete, can someone complete it please) Edited by kc5170
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.