Difference between revisions of "2022 AMC 10B Problems/Problem 16"
Pi is 3.14 (talk | contribs) |
Kavya.rajesh (talk | contribs) (Problem 3) |
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+ | ==Solution 3 (Area of trapezoid)== | ||
+ | Proceed similar to solution 1 and use similar triangles to find side length of GE. Then use area of a trapezoid to solve for the area of ACEG. | ||
== Video Solution by OmegaLearn Using Similar Triangles == | == Video Solution by OmegaLearn Using Similar Triangles == |
Revision as of 06:04, 18 November 2022
- The following problem is from both the 2022 AMC 10B #16 and 2022 AMC 12B #13, so both problems redirect to this page.
Contents
Problem
The diagram below shows a rectangle with side lengths 4 and 8 and a square with side length 5. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
Solution 1
Let us label the points on the diagram.
By doing some angle chasing using the fact that and
are right angles, we find that
. Similarly,
. Therefore,
.
As we are given a rectangle and a square, and
. Therefore,
is a 3-4-5 right triangle and
.
is also
. So, using the similar triangles,
and
.
. Using the similar triangles again,
is
of the corresponding
. So,
Finally, we have
~Connor132435
Solution 2 (Clever)
(Refer to the diagram above)
Proceed the same way as solution 1 until you get all of the side lengths. Then, it is clear that due to the answer choices, we only need to find the fractional part of the shaded area. The area of the whole rectangle is integral, as is the area of ,
, and the rectangle to the far left of the diagram. The area of
is
and thus the fractional part of the answer is
. Our answer is
~mathboy100
Solution 3 (Area of trapezoid)
Proceed similar to solution 1 and use similar triangles to find side length of GE. Then use area of a trapezoid to solve for the area of ACEG.
Video Solution by OmegaLearn Using Similar Triangles
~ pi_is_3.14
See Also
2022 AMC 10B (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 AMC 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.