Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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==Solution 3== | ==Solution 3== | ||
Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>. | Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>. | ||
− | Let the base of the small triangles of area 1 be <math>x</math>, then the base length of <math>\Delta ADE=4x</math>. Notice, <math>\ | + | Let the base of the small triangles of area 1 be <math>x</math>, then the base length of <math>\Delta ADE=4x</math>. Notice, <math>\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math> |
− | Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\ | + | Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.</math> |
<asy> | <asy> | ||
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draw((15,15)--(45,15)); | draw((15,15)--(45,15)); | ||
</asy> | </asy> | ||
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==Solution 4== | ==Solution 4== |
Revision as of 01:32, 20 January 2021
- The following problem is from both the 2018 AMC 12A #8 and 2018 AMC 10A #9, so both problems redirect to this page.
Contents
Problem
All of the triangles in the diagram below are similar to isosceles triangle , in which
. Each of the 7 smallest triangles has area 1, and
has area 40. What is the area of trapezoid
?
Solution 1
Let be the area of
. Note that
is comprised of the
small isosceles triangles and a triangle similar to
with side length ratio
(so an area ratio of
). Thus, we have
This gives
, so the area of
.
Solution 2
Let the base length of the small triangle be . Then, there is a triangle
encompassing the 7 small triangles and sharing the top angle with a base length of
. Because the area is proportional to the square of the side, let the base
be
. Then triangle
has an area of 16. So the area is
.
Solution 3
Notice .
Let the base of the small triangles of area 1 be
, then the base length of
. Notice,
, then
Thus,
Solution 4
The area of is 16 times the area of the small triangle, as they are similar and their side ratio is
. Therefore the area of the trapezoid is
.
Solution 5
You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be , so to find the area of such trapezoid
, we just take
, like so.
Solution 6
The combined area of the small triangles is , and from the fact that each small triangle has an area of
, we can deduce that the larger triangle above has an area of
(as the sides of the triangles are in a proportion of
, so will their areas have a proportion that is the square of the proportion of their sides, or
). Thus, the combined area of the top triangle and the trapezoid immediately below is
. The area of trapezoid
is thus the area of triangle
.
Solution 7
You can assume for the base of one of the smaller triangles to be and the height to be
, giving an area of 1. The larger triangle above the 7 smaller ones then has base
and height
, giving it an area of
. Then the area of triangle
is
and
.
Video Solution
~savannahsolver
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.