Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 241: 23

Base =8 Altitude= 6 hypotenuse 10

By Pythagoras theorem $Hypotenuse^{2}= Base^{2}+ Altitude^{2}$ $(2x)^{2} = (x+3)^{2}+(x+1)^{2}$ $4x^{2} = x^{2}+9+6x+x^{2}+1+2x$ $2x^{2}-8x-10=0$ $x^{2}-4x-5=0$ $x^{2}-(5-1)x-5=0$ $x^{2}-5x+x-5=0$ $(x-5)(x+1)=0$ so x=5 is only valid value of x Base= (x+3)=5+3=8 Altitude=(x+1)=5+1=6 hypotenuse=2x=10