Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - Chapter Review - Page 424: 38

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 $A(-1, 1)$, $B(-1, -2)$, and $C(3, -2)$. 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: $AB = \sqrt {(-1 - (-1))^2 + (-2 - 1)^2}$ Simplify within the parentheses: $AB = \sqrt {(0)^2 + (-3)^2}$ Evaluate the exponents: $AB = \sqrt {0 + 9}$ Add what is underneath the radical: $AB = \sqrt {9}$ Take the square root of $9$ to simplify: $AB = 3$ Let's look at the next side, $BC$: $BC = \sqrt {(3 - (-1))^2 + (-2 - (-2))^2}$ Simplify within the parentheses: $BC = \sqrt {(4)^2 + (0)^2}$ Evaluate the exponents: $BC = \sqrt {16 + 0}$ Add what is underneath the radical: $BC = \sqrt {16}$ Take the square root of $16$ to simplify: $BC = 4$ Let's look at $CA$: $CA = \sqrt {(3 - (-1))^2 + (-2 - 1)^2}$ Simplify within the parentheses: $CA = \sqrt {(4)^2 + (-3)^2}$ Evaluate the exponents: $CA = \sqrt {16 + 9}$ Add what is underneath the radical: $CA = \sqrt {25}$ Take the square root of $25$: $CA = 5$ All sides are of different lengths; therefore, this triangle is scalene.
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