Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 10 - Area - 10-1 Areas of Parallelograms and Triangles - Practice and Problem-Solving Exercises: 44

Answer

34 in$^{2}$

Work Step by Step

A = $\sqrt {s(s-a)(s-b)(s-c)}$ where s = $\frac{1}{2}$(a+b+c) a = 8 b = 9 c = 10 To simplify the equation and make the computation process easier, solve for "s" first. s = $\frac{1}{2}$(a+b+c) = $\frac{1}{2}$(8+9+10) = $\frac{1}{2}$(27) = 13.5 Now, substitute 13.5 into the original equation every time "s" appears. A = $\sqrt {s(s-a)(s-b)(s-c)}$ A = $\sqrt {13.5(13.5-a)(13.5-b)(13.5-c)}$ A = $\sqrt {13.5(13.5-8)(13.5-9)(13.5-10)}$ A = $\sqrt {13.5(5.5)(4.5)(3.5)}$ A = $\sqrt {1169.4375}$ A $\approx$ 34.2 34.2 rounded to the nearest whole number is 34.
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