Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 12 - Section 12.5 - Circles - Exercise - Page 411: 32

Answer

a) $A=56.9~ft^{2}$ b) $c=4.16ft$ or $49.9~in$

Work Step by Step

(a) The area of the workstation is the area of the square minus the area of the circle formed by the 4 corner segments. We use the formula $A=a^{2}-\pi r^{2}$ Where $a=8ft$ $r=18in=1.5ft$ ($1~foot=12~inches$) So, $A=(8ft)^{2}-\pi (1.5ft)^{2}=56.93ft^{2}\approx56.9~ft^2$ (b) The distance from the corner to the center of the workstation ($c$) is half the hypotenuse of the triangle formed by the sides of the table minus the radius of the corner. Using the Pythagorean Theorem, we get: $c=\frac{1}{2}\sqrt {(8ft)^{2}+(8ft)^{2}}-1.5ft=4.16ft$ We can convert this to inches as well ($12~inches=1~ft)$: $4.16*12~inches=49.9~inches$
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