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> | 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 18:56, 24 January 2023
By AA~ similarity triangle we can find the ratio of the area of big: small —>then there are a relative
for the
trapezoids combines. For
trapezoid it is a relative
so now the ratio is
which can simplify to
Animated Video Solution
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Written Solution
By AA~ similarity triangle we can find the ratio of the area of big: small —> 
then there are a relative 
for the 
trapezoids combines. For 
trapezoid it is a relative 
so now the ratio is 
which can simplify to 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
