Difference between revisions of "2022 AMC 10B Problems/Problem 2"
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Since we know that <math>\triangle ABD</math> ≅ <math>\triangle BDC</math> and that <math>ABCD=\triangle ABD + \triangle BDC</math>. That means we double the area of <math>\triangle ADB</math>, <math>10 \cdot 2 = \boxed{\textbf{(D) }20}</math>. | Since we know that <math>\triangle ABD</math> ≅ <math>\triangle BDC</math> and that <math>ABCD=\triangle ABD + \triangle BDC</math>. That means we double the area of <math>\triangle ADB</math>, <math>10 \cdot 2 = \boxed{\textbf{(D) }20}</math>. | ||
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== See Also == | == See Also == |
Revision as of 01:33, 18 November 2022
- The following problem is from both the 2022 AMC 10B #2 and 2022 AMC 12B #2, so both problems redirect to this page.
Contents
Problem
In rhombus , point lies on segment so that , , and . What is the area of ? (Note: The figure is not drawn to scale.)
Solution
(Figure redrawn to scale.)
.
is a rhombus, so .
is a 3-4-5 right triangle, so .
The area of the rhombus .
~richiedelgado
Solution 2
The diagram is the same as solution 1, just constrcuted a line at
When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. Which means ≅ by the Alternate Interior Angles Theorem.
By knowing these statements, we get ≅ .
Knowing that ; , then we would know that because is a 3-4-5 right triangle (as stated in solution 1).
We get the area of as 10 because of the area of the triangle > .
Since we know that ≅ and that . That means we double the area of , .
-ghfhgvghj10
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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.