Two equal sides of a triangle are each 4 m
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 $23 m$.

 

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