Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-7 Polygons in the Coordinate Plane - Practice and Problem-Solving Exercises - Page 403: 5

Answer

scalene

Work Step by Step

We use the distance formula to determine what type of triangle is pictured. The vertices of the triangle are $A(-2, 3)$, $B(2, 2)$, and $C(-2, -1)$. The distance formula is given by the following formula: $d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Let's determine the lengths of the different sides of the triangle. We'll look at $AB$ first: $AB = \sqrt {(2 - (-2))^2 + (2 - 3)^2}$ Simplify within the parentheses: $AB = \sqrt {(4)^2 + (-1)^2}$ Evaluate the exponents: $AB = \sqrt {16 + 1}$ Add what is underneath the radical: $AB = \sqrt {17}$ Let's look at the next side, $BC$: $BC = \sqrt {(-2 - 2)^2 + (-1 - 2))^2}$ Simplify within the parentheses: $BC = \sqrt {(-4)^2 + (-3)^2}$ Evaluate the exponents: $BC = \sqrt {16 + 9}$ Add what is underneath the radical: $BC = \sqrt {25}$ Evaluate to solve: $BC = 5$ Let's look at $CA$: $CA = \sqrt {(-2 - (-2))^2 + (-1 - 3)^2}$ Simplify within the parentheses: $CA = \sqrt {(0)^2 + (-4)^2}$ Evaluate the exponents: $CA = \sqrt {0 + 16}$ Add what is underneath the radical: $CA = \sqrt {16}$ Evaluate to solve: $CA = 4$ All three sides have different lengths; therefore, this triangle is scalene.
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