If the radius of a sector is 14cm and \$\theta = 90^0\$. Find the perimeter

Arc length of a sector.

l = \$\frac{\theta}{360^\circ} \times 2 \times \pi r\$

Substitute \$22/7\$ for \$pi\$, \$90^\circ\$ for \$\theta\$ and 14 for r.

l = \$\frac{90°}{360°} \times 2 \times \frac{22}{7} \times 14\$

Cancel out common factor.

\$l = \frac{\cancel{90°}}{\cancel{360°}} \times 2 \times \frac{22}{\cancel{7}^1} \times \cancel{14}^2 \$

Cancel out common factor.

\$l = \frac{1}{\cancel{4}^2} \times \cancel{2}^1 \times \frac{22}{1} \times 2 \$

Cancel out common factor.

\$l = \frac{1}{\cancel{2}^1} \times 1 \times\frac{\cancel {22} ^{11}}{1} \times 2 \$

Multiply.

\$l = 22 \$

Perimeter of a sector.

\$P = 2r + l \$

Substitute 22 for l and 14 for r.

\$ P = 2(14) + 22 \$

Multiply.

\$P = 28 + 22 \$

Add.

\$ P = 50 \$

Answer

The perimeter is \$50 cm\$.