Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 371: 46b

Answer

Not

Work Step by Step

(8,10,12) Using Heron's formula if three sides of a triangle have length a,b,c then area A of a triangle is A = $\sqrt s(s-a)(s-b)(s-c)$ where s = $\frac{a+b+c}{2}$ Lets take a=8 b=10 and c=12 s = $\frac{8+10+12}{2}$ = 15 A = $\sqrt 15(15-8)(15-10)(15-12)$ =$\sqrt 15 * 7 * 5 * 3$ =15$\sqrt 7$ $unit^{2}$ =39.686 $unit^{2}$ If a,b,c be the integer length of the sides of the triangle. If area of the triangle is also an integer than (a,b,c) is known as heron triplet. Since area is not integer i.e. 39.686 so (8,10,12) is not a heron triples.
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