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: 37

Answer

8 in.

Work Step by Step

The area of $\triangle$ ABC is 40$in^{2}$ $\angle$C = 90$^{\circ}$, AC= x, BC = x+2 We need to find out the value of x The area of right angle triangle with legs of length a and b is given by A = $\frac{1}{2}$ab From given right triangle ABC Lengths of legs a= x+2 b=x Area of $\triangle$ ABC = $\frac{1}{2}$ab 40 $in^{2}$ = $\frac{1}{2}$x*(x+2) 80 = $x^{2}$ + 2x $x^{2}$ + 2x - 80 = 0 $x^{2}$ + 10x - 8x - 80 = 0 x(x+10) - 8(x+10) = 0 (x-8)(x+10) = 0 x =8,-10 Therefore the value of x is 8in. because it cant be negative.
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.