Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 324: 71


The triangle has whole number lengths and its area and perimeter is both 36. Therefore, it is a perfect triangle.

Work Step by Step

A triangle is a perfect triangle if it fulfills two requirements. First, a perfect triangle is a triangle whose sides have whole number lengths. Since the triangle in question has sides 9,10 and 17; it fulfills the first requirement. Second, the area of a perfect triangle is numerically equal to its perimeter. Therefore, we find the perimeter and area of the triangle in question and compare the two: Perimeter= Sum of all three sides Perimeter=$9+10+17$ Perimeter=$36$ To find the area, we need to apply the Heron's formula since the length of all three sides is given: To use the Heron's formula, we first need to find the semi-perimeter $s$. We substitute the values of the sides of the triangle in the formula below to find the semi-perimeter $s$: $s=\frac{1}{2}(a+b+c)$ $s=\frac{1}{2}(9+10+17)$ $s=\frac{1}{2}(36)$ $s=18$ Now, we use the Heron's formula to find the area of the triangle: $A=\sqrt{s(s-a)(s-b)(s-c)}$ $A=\sqrt{18(18-9)(18-10)(18-17)}$ $A=\sqrt{18(9)(8)(1)}$ $A=\sqrt{1296}$ $A=36$ Since the area of the triangle is numerically equal to its perimeter, the triangle also fulfills the second requirement. Therefore, it is a perfect triangle.
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.