Difference between revisions of "2022 AMC 10B Problems/Problem 16"
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<math>4=-\frac{4}{3}x+12 \frac{1}{3}</math>, so <math>x=6\frac{1}{4}</math> and <math>y=4</math> This would mean <math>G\left(6\frac{1}{4},4\right)</math>. | <math>4=-\frac{4}{3}x+12 \frac{1}{3}</math>, so <math>x=6\frac{1}{4}</math> and <math>y=4</math> This would mean <math>G\left(6\frac{1}{4},4\right)</math>. | ||
− | Since we have our <math>G</math> coordinate, we can continue with Solution 3, with the area of the trapezoid <math>\left(\frac{ | + | Since we have our <math>G</math> coordinate, we can continue with Solution 3, with the area of the trapezoid <math>\left(\frac{EG+AC}{2}\right)(CE)</math>, where <math>EG=\frac{5}{4}</math> (using distance formula for <math>E</math> to <math>G</math>), <math>AC=5</math>, and <math>CE=5</math>. |
By substitution, we get <math>\left(\frac{\frac{5}{4}+5}{2}\right)(5)=</math><math>\boxed{\textbf{(D) }15\dfrac{5}{8}}</math>. | By substitution, we get <math>\left(\frac{\frac{5}{4}+5}{2}\right)(5)=</math><math>\boxed{\textbf{(D) }15\dfrac{5}{8}}</math>. |
Revision as of 16:15, 14 December 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 and
and a square with side length
. 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
-
-
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 (Coordinate Geometry)
Same diagram as Solution 1, but added point , which is
. I also renamed all the points to form coordinates using
as the origin.
In order to find the area, point 's coordinates must be found. Notice how
and
intercept at point
. This means that we need to find the equations for
and
and make a system of linear equations.
Using the slope formula , we get the slope for
, which means
Then, by using point-slope form. . We can say that the equation for
is
or in this case,
.
And it is easy to figure out that the equation for is
.
The best way to solve the system of linear equations is to substitute the for the
in equation
.
, so
and
This would mean
.
Since we have our coordinate, we can continue with Solution 3, with the area of the trapezoid
, where
(using distance formula for
to
),
, and
.
By substitution, we get .
~ghfhgvghj10 (+ minor edits ~TaeKim)
Video Solution 1
~Education, the Study of Everything
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.