Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1077: 112


$b=3\ ,l=\ 8$.

Work Step by Step

We have the formula for the perimeter: $P=2\left( l+b \right)$ So $\begin{align} & 22=2\left( l+b \right) \\ & 11=l+b \\ & l=11-b \end{align}$ The area is $A=l\times b$ , So, $24=l\times b$ Substituting the value of the length in equation $24=l\times b$, we get ${{b}^{2}}-11b+24=0$ On solving the above quadratic equation, we get $b=3\text{ and}\ 8$. Substituting $b$ in equation $l=11-b$. When $b=3$ the value of $l=8$ and when $b=8$ the value of $l=3$.
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