Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | By similarity, the height is <math>3+\ | + | By similarity, the height is <math>3+\frac31\cdot2=9</math> and the base is <math>\frac92\cdot1=4.5</math>. |
Thus the area is <math>\frac{9\cdot4.5}2=20.25=20\frac14</math>, or <math>\boxed{(\textbf{B})}</math>. | Thus the area is <math>\frac{9\cdot4.5}2=20.25=20\frac14</math>, or <math>\boxed{(\textbf{B})}</math>. | ||
~Hefei417, or 陆畅 Sunny from China | ~Hefei417, or 陆畅 Sunny from China | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:05, 23 November 2021
Problem
IDK Don't ask me.
Solution
By similarity, the height is 
and the base is 
.
Thus the area is 
, or 
.
~Hefei417, or 陆畅 Sunny from China
See Also
| 2021 Fall AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
