Basic College Mathematics (10th Edition)

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

Chapter 8 - Geometry - 8.3 Rectangles and Squares - 8.3 Exercises - Page 553: 36


(a) Perimeter = $30~feet$, Area= $54~feet^2$ (b) Perimeter multiplied by 3. Area multiplied by 9. (c) Perimeter multiplied by 4. Area multiplied by 16.

Work Step by Step

(a) We know the the dimensions are now tripled: length: $3*3=9$ feet width: $2*3=6$ feet (See sketch.) New perimeter: $9*2+6*2=18+12=30$ feet New area: $9*6=54~feet^2$ (b) We see that the perimeter has tripled: $\displaystyle \frac{new~perimeter}{old~perimeter}=\frac{30}{10}=3$ We see that the area is 9 times the original area: $\displaystyle \frac{new~area}{old~area}=\frac{54}{6}=9$ (c) If the length and width were multiplied by $4$, we would expect the perimeter to multiply by $4$ and the area to multiply by $16$. This is because the perimeter multiplies by the same factor that the lengths do and the area multiplies by that factor squared ($4^2=16$). This is a general rule of geometry.
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