Question

The perimeter of a triangle is 57 inches. Twice the length of the longest side minus the length of the shortest side is 22 inches. The sum of the length of the longest side and twice the sum of both the other side lengths is 94 inches. Find the side lengths

Answer 1

`\begin{cases} a = shortest\\ b = medium\\ c = longest \end{cases} \begin{array}{llll} \stackrel{\textit{perimeter is 57}~\hfill }{a + b + c = 57}~\\\\\stackrel{\textit{twice longest minus shortest}}{2c-a=22~\hfill }\\\\ \stackrel{\textit{longest plus twice others}}{c + 2(a+b) = 94~\hfill } \end{array} \\\\[-0.35em] ~\dotfill\\\\ 2c-a=22\implies 2c=a+22\implies \boxed{2c-22=a} \\\\\\ \stackrel{\textit{we know that}}{c+2(a+b)=94}\implies c+2a+2b=94\implies c+2(2c-22)+b=94`

`c+4c-44+2b=94\implies 5c-44+2b=94\implies 5c+2b=138 \\\\\\ 2b=138-5c\implies \boxed{b = \cfrac{138-5c}{2}} \\\\\\ \stackrel{\textit{we know the perimeter is}}{57=a + b + c}\implies 57 = \stackrel{a}{(2c-22)}+\stackrel{b}{\cfrac{138-5c}{2}}+c \\\\\\ 57=2c-22+\cfrac{138}{2}-\cfrac{5c}{2}+c\implies 57=3c-22+69-\cfrac{5c}{2} \\\\\\ 57=3c-47+\cfrac{5c}{2}\implies 10=3c-\cfrac{5c}{2}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{20=6c-5c}`

`\blacktriangleright 20=c \blacktriangleleft \\\\\\ \boxed{2c-22=a}\implies 40-22=a\implies \blacktriangleright 18=a \blacktriangleleft \\\\\\ \boxed{b = \cfrac{138-5c}{2}}\implies b=\cfrac{138-5(20)}{2}\implies b=\cfrac{38}{2}\implies \blacktriangleright 19 \blacktriangleleft`