## Elementary Geometry for College Students (7th Edition)

Published by Cengage

# Chapter 8 - Review Exercises - Page 395: 2

#### Answer

a) 40 b) $40\sqrt3$ c) $40\sqrt{2}$

#### Work Step by Step

In each case, we must find the height to find the area. a) We know this forms a 30-60-90 triangle, so the height it: $h = 1/2(10) = 5$ $A = (b)(h) = (8)(5) = 40$ b) We know this is a 30-60-90 triangle, but this time A is the 60 degree angle. Thus: $h = 10 \times \frac{\sqrt3}{2}= 5\sqrt3$ $A = (b)(h) = (5\sqrt3)(8) = 40\sqrt3$ c) The third triangle formed is a 45-45-90 triangle. Thus: $h = 10 \times \frac{\sqrt{2}}{2} = 5 \sqrt2$ $A = (5\sqrt{2})(8) = 40\sqrt{2}$

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