Question:
Two equal sides of a triangle are each 4 m less than three times the third side. Find the dimensions of the triangle, if its perimeter is 55 m.
Solution:
Let the third side of triangle be $x \mathrm{~m}$.
Then, two equal sides of triangle $=(3 x-4) \mathrm{m}$
Given, perimeter of triangle $=55 \mathrm{~m}$
$\because$ Perimeter of a triangle $\simeq$ Sum of the sides of the triangle
$\therefore \quad x+3 x-4+3 x-4=55$
$\Rightarrow \quad 7 x-8=55 \Rightarrow 7 x=55+8$
$\Rightarrow$ $7 x=63 \Rightarrow x=\frac{63}{7}$
$\therefore \quad x=9$
Hence, the dimensions of the triangle are $9 \mathrm{~m},(3 \times 9-4)$ mand $(3 \times 9-4) \mathrm{m}$, i.e. $9,(27-4)$, and $(27-4) m$, i.e. $9 m, 23 m$ and $23 m$.