Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 350: 37

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.

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