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: 45b

Answer

yes

Work Step by Step

(13,14,15) 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=13 b=14 and c=15 s = $\frac{13+14+15}{2}$ = 21 A = $\sqrt 21(21-13)(21-14)(21-15)$ =$\sqrt 21*8*7*6$ = 84 $unit^{2}$ If a,b,c be the integer length of the sides of a triangle . If the area of the triangle is also a integer then (a,b,c) is known as a heron triplet. Since A = 84 is a integer (a,b,c) = (13,14,15) integer Therefore the triplet is a heron triplet.
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