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 8.1 - Area and Initial Postulates - Exercises - Page 360: 38

Answer

6,8

Work Step by Step

Let us assume the legs of a right triangle is 2n, 2+2 The given condition is the numerical value of the area of triangle is three times of that the longer leg. Area of triangle = 3(2n+2) The area of the right triangle with legs of the length a and b given by A = $\frac{1}{2}$ ab 3(2n+2)=$\frac{1}{2}$ 2n (2n+2) 6n+6 = n(2n+2) 6n+6 = 2$n^{2}$ + 2n 2$n^{2}$ - 4n - 6 =0 $n^{2}$ - 2n - 3 =0 $n^{2}$ - 3n + n - 3 =0 n(n-3)+1(n-3) = 0 (n+1)(n-3) = 0 n=-1,3 Therefore the n value cannot be negative, the value of n = 3 Therefore the legs of triangle 2n , 2n+2 = 2*3, 2*3+2 =6,8 6 and 8 are the length of the required triangle.
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