Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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− | All of the triangles in the diagram below are similar to | + | {{duplicate|[[2018 AMC 10A Problems/Problem 9|2018 AMC 10A #9]] and [[2018 AMC 12A Problems/Problem 8|2018 AMC 12A #8]]}} |
+ | |||
+ | ==Problem== | ||
+ | |||
+ | All of the triangles in the diagram below are similar to isosceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each of the <math>7</math> smallest triangles has area <math>1,</math> and <math>\triangle ABC</math> has area <math>40</math>. What is the area of trapezoid <math>DBCE</math>? | ||
<asy> | <asy> | ||
Line 22: | Line 26: | ||
==Solution 1== | ==Solution 1== | ||
− | + | Let <math>x</math> be the area of <math>ADE</math>. Note that <math>x</math> is comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area ratio of <math>9:16</math>). Thus, we have <cmath>x=7+\dfrac{9}{16}x.</cmath> This gives <math>x=16</math>, so the area of <math>DBCE=40-x=\boxed{(E) 24}</math>. | |
==Solution 2== | ==Solution 2== | ||
− | Let <math>x</math> | + | Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. The ratio of the area of triangle <math>ADE</math> to triangle <math>ABC</math> is <math>\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}</math>. The problem says the area of triangle <math>ABC</math> is <math>40</math>, so the area of triangle <math>ADE</math> is <math>16</math>. So the area of trapezoid <math>DBCE</math> is <math>40 - 16 = \boxed{24}</math>. |
==Solution 3== | ==Solution 3== | ||
+ | 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>\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]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.</math> | ||
+ | |||
+ | <asy> | ||
+ | unitsize(5); | ||
+ | dot((0,0)); | ||
+ | dot((60,0)); | ||
+ | dot((50,10)); | ||
+ | dot((10,10)); | ||
+ | dot((30,30)); | ||
+ | draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); | ||
+ | draw((10,10)--(50,10)); | ||
+ | label("$B$",(0,0),SW); | ||
+ | label("$C$",(60,0),SE); | ||
+ | label("$E$",(50,10),E); | ||
+ | label("$D$",(10,10),W); | ||
+ | label("$A$",(30,30),N); | ||
+ | draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); | ||
+ | draw((15,15)--(45,15)); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 4== | ||
The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>. | The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>. | ||
− | == See Also == | + | ==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 <math>7+5+3+1=16</math>, so to find the area of such trapezoid <math>BCED</math>, we just take <math>40-16=\boxed{24}</math>, like so. | ||
+ | |||
+ | ==Solution 6== | ||
+ | The combined area of the small triangles is <math>7</math>, and from the fact that each small triangle has an area of <math>1</math>, we can deduce that the larger triangle above has an area of <math>9</math> (as the sides of the triangles are in a proportion of <math>\frac{1}{3}</math>, so will their areas have a proportion that is the square of the proportion of their sides, or <math>\frac {1}{9}</math>). Thus, the combined area of the top triangle and the trapezoid immediately below is <math>7 + 9 = 16</math>. The area of trapezoid <math>BCED</math> is thus the area of triangle <math>ABC-16 =\boxed{24}</math>. | ||
+ | |||
+ | ==Solution 7== | ||
+ | You can assume for the base of one of the smaller triangles to be <math>\frac{1}{a}</math> and the height to be <math>2a</math>, giving an area of 1. The larger triangle above the 7 smaller ones then has base <math>\frac{3}{a}</math> and height <math>6a</math>, giving it an area of <math>9</math>. Then the area of triangle <math>ADE</math> is <math>16</math> and <math>40-16=\boxed{24}</math>. | ||
+ | |||
+ | ==Solution 8== | ||
+ | You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence <math>7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)</math> . Using the infinite geometric series sum formula gives us <math>7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16</math>. The triangle's area would thus be <math>40-16=\boxed{24}</math>. | ||
+ | |||
+ | -ConcaveTriangle | ||
+ | |||
+ | ==Solution 9 (weird)== | ||
+ | Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is <math>1</math>, and thus its base is <math>2.</math> | ||
+ | |||
+ | Let <math>h</math> be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and <math>\triangle{ADE}</math>. The top triangle has a base of <math>3\cdot{2}=6</math>, and <math>DE=4\cdot{2}=8.</math> The height of <math>\triangle{ADE}</math> is <math>h+1</math>, therefore our ratio is <math>\frac{h}{6}=\frac{h+1}{8}</math>, which yields <math>h=3</math> as our answer. | ||
+ | |||
+ | To find the area of the trapezoid, we can take the area of <math>\triangle{ABC}</math> and subtract the area of <math>\triangle{ADE},</math> whose base is <math>8</math> and height <math>3+1=4</math>. It follows that the area of <math>\triangle{ADE}=16</math>, and subtracting this from <math>40</math> gives us <math>40-16=\boxed{\text{D) }24}</math>. | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!)== | ||
+ | https://youtu.be/lgkkomUwSs0 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solutions== | ||
+ | |||
+ | https://youtu.be/HJALwsbHZXc | ||
+ | |||
+ | ~Whiz | ||
+ | |||
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/tq4rWskQDh8 | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | https://youtu.be/4_x1sgcQCp4?t=2959 | ||
+ | |||
+ | ~pi_is_3.14 | ||
+ | |||
+ | ==See Also== | ||
+ | |||
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}} | ||
{{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 14:44, 3 July 2023
- The following problem is from both the 2018 AMC 10A #9 and 2018 AMC 12A #8, 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
smallest triangles has area
and
has area
. 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
. The ratio of the area of triangle
to triangle
is
. The problem says the area of triangle
is
, so the area of triangle
is
. So the area of trapezoid
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
.
Solution 8
You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence . Using the infinite geometric series sum formula gives us
. The triangle's area would thus be
.
-ConcaveTriangle
Solution 9 (weird)
Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is , and thus its base is
Let be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and
. The top triangle has a base of
, and
The height of
is
, therefore our ratio is
, which yields
as our answer.
To find the area of the trapezoid, we can take the area of and subtract the area of
whose base is
and height
. It follows that the area of
, and subtracting this from
gives us
.
-Benedict T (countmath1)
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solutions
~Whiz
~IceMatrix
~savannahsolver
https://youtu.be/4_x1sgcQCp4?t=2959
~pi_is_3.14
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.