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2023 AMC 8 Problems/Problem 19: Difference between revisions

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By AA~ similarity triangle we can find the ratio of the area of big: small —> <math>\frac{9}{4}</math> then there are a relative <math>5</math> for the <math>3</math> trapezoids combines. For <math>1</math> trapezoid it is a relative <math>53</math> so now the ratio is <math>\frac{5}{\frac{3}{4}}</math> which can simplify to <math>\boxed{\text(C) \frac{5}{12}}</math>
 
==Written Solution==
 
By AA~ similarity triangle we can find the ratio of the area of big: small —> <math>\frac{9}{4}</math> then there are a relative <math>5</math> for the <math>3</math> trapezoids combines. For <math>1</math> trapezoid it is a relative <math>53</math> so now the ratio is <math>\frac{5}{\frac{3}{4}}</math> which can simplify to <math>\boxed{\text(C) \frac{5}{12}}</math>
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 




Line 6: Line 12:


~Star League (https://starleague.us)
~Star League (https://starleague.us)
==Written Solution==
By AA~ similarity triangle we can find the ratio of the area of big: small —> <math>\frac{9}{4}</math> then there are a relative <math>5</math> for the <math>3</math> trapezoids combines. For <math>1</math> trapezoid it is a relative <math>53</math> so now the ratio is <math>\frac{5}{\frac{3}{4}}</math> which can simplify to <math>\boxed{\text(C) \frac{5}{12}}</math>
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Revision as of 19:59, 24 January 2023

Written Solution

By AA~ similarity triangle we can find the ratio of the area of big: small —> $\frac{9}{4}$ then there are a relative $5$ for the $3$ trapezoids combines. For $1$ trapezoid it is a relative $53$ so now the ratio is $\frac{5}{\frac{3}{4}}$ which can simplify to $\boxed{\text(C) \frac{5}{12}}$

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat


Animated Video Solution

https://youtu.be/Xq4LdJJtbDk

~Star League (https://starleague.us)

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