Elementary Geometry for College Students (7th Edition)

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



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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.