Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 94

See below.

By Pythagorean's Theorem we know that $a^2+b^2=c^2$, where $a$ and $b$ are the legs, $c$ is the hypotenuse. Hence here $x^2+(3x+13)^2=(4x+5)^2\\x^2+9x^2+78x+169=16x^2+40x+25\\38x+144=6x^2\\3x^2-19x-72=0\\(3x+8)(x-9)=0$ Thus $x=-8/3$ or $x=9$ but $x$ cannot be negative, so $x=9$, thus there is one triangle with the dimensions of $9,40,41$