Difference between revisions of "2023 AMC 8 Problems/Problem 19"
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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> | ||
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+ | ~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
~Star League (https://starleague.us)
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