Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 102: 41

Answer

(a) $AC^2=AB^2+BC^2$ $50=10+40$ $50=50$ (b) $Area=10$

Work Step by Step

(a) According to the definition of Pythagorean Theorem, square of the biggest side of a triangle equals to the sum of squares of the other sides. Let's first calculate length of each side using distance formula: $AB=\sqrt{(3-2)^2+(-1-2)^2}=\sqrt{1+9}=\sqrt{10}$ $BC=\sqrt{(-3-3)^2+(-3+1)^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}$ $AC=\sqrt{(-3-2)^2+(-3-2)^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$ As we see, $AC$ is the biggest one, so to show whether the triangle is right or not we can write: $AC^2=AB^2+BC^2$ $50=10+40$ $50=50$ (b) Area of triangle is $Height \times Base \times \frac{1}{2}$. In case of right triangle, Height and Base are $adjacent$ and $opposite$: $A=\sqrt{10}\times 2\sqrt{10} \times \frac{1}{2}=10$
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