Chapter 14.2, Problem 62STP

To determine

To Find: Length of all three sides of triangle.

Answer to Problem 62STP

(C)

Explanation of Solution

Given information :

Perimeter of a right-angled triangle 36 inches. Twice the length of longer leg minus the length of

shorter leg is 6 inches

  $\begin{array}{l}AB+BC+CA=36\text{ longer leg, BC, Shorter leg AB}\\ 2BC-2AB=6\Rightarrow BC-AB=3\\ c+a+b=36\mathrm{.....}(1)\\ a-c=3\\ a=3+c\mathrm{......}(2)\\ Puttingin(1)\\ c+3+c+b=36\\ 2c+3+b=36\\ 2c+b=33\\ b=33-2c\mathrm{.....}(3)\\ \text{Using Pythagoras Theorem,}\\ A{C}^2=A{B}^2+B{C}^2\\ b^2=c^2+a^2\\ \text{From (2) and (3)}\\ $ {(33-2c)}^2=c^2+{(3+c)}^2$1089+4c^2-132c=c^2+9+c^2+6c\\ 2c^2-138c+1080=0\\ c^2-69c+540=0\\ c^2-60c+9c+540=0\\ c(c-60)-9(c-60)=0\\ (c-60)(c-9)=0\\ c=60andc=9\\ c=60,\text{ is not possible, since perimter is 36 inches}\\ \text{AB = 9 inch}\\ \text{BC = 9 + 3 = 12 inch}\\ \text{AC}=\sqrt{{12}^2+9^2}=\sqrt{144+81}=\sqrt{225}=15\text{ inch, Option (c)}\end{array}$