Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10: 0134467795
ISBN 13: 978-0-13446-779-5

Chapter 8 - Geometry - 8.9 Similar Triangles - 8.9 Exercises - Page 600: 12


$\displaystyle \frac{AB}{PQ}=\frac{2}{3}$ $\displaystyle \frac{AC}{PR}=\frac{2}{3}$ $\displaystyle \frac{BC}{QR}=\frac{2}{3}$

Work Step by Step

We find the ratio of sides between the two triangles and simplify: $\displaystyle \frac{AB}{PQ}=\frac{22~cm}{33~cm}=\frac{2*11}{3*11}=\frac{2}{3}$ $\displaystyle \frac{AC}{PR}=\frac{30~cm}{45~cm}=\frac{2*15}{3*15}=\frac{2}{3}$ $\displaystyle \frac{BC}{QR}=\frac{16~cm}{24~cm}=\frac{2*8}{3*8}=\frac{2}{3}$ The ratios are all the same, which makes sense for proportional triangles.
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