The perimeter of a sector is 36m and \$\theta = 180^0\$. Find the radius?

Perimeter of a sector

\begin{align} P &= 2r + l \\ P &= 2r + \frac{\theta}{360^\circ} \times 2 \times \pi \times r \\ P &= \biggl[ 2 + \frac{\theta}{360^\circ} \times 2 \times \pi \biggr] \times r \\ \end{align}

Substitute \$180°\$ for \$\theta\$ and 36 for P, and \$22/7\$ for \$\pi\$.

\$36 = (2 + \frac{180°}{360°} \times 2 \times \frac{22}{7}) \times r \$

Cancel out common factor.

\$36 = (2 + \frac{\cancel{180°} ^{2}}{\cancel{360°} ^{4}} \times 2 \times \frac{22}{7}) \times r \$.

Cancel out common factor.

\$ 36 = (2 + \frac{\cancel {2}^{1}}{\cancel{4}^{2}} \times 2 \times \frac{22}{7}) \times r \$

Cancel out common factor.

\$36 = (2 + \frac{1}{\cancel{2}^{1}} \times \cancel{2}^1 \times \frac{22}{7}) \times r \$

Cancel out common factor.

\$36 = (2 + \frac{22}{7}) \times r \$

Add.

\$36 = ( \frac{36}{7}) \times r \$

Cancel out common factor.

\$ \cancel{36}^1 = (\frac{\cancel{36}^1}{7}) \times r \$

Solve for r.

\$r = 7 \$

Answer

The radius is \$7 m\$.