Question

If the perimeter of a semicircular protractor is $36\mathrm{cm}$, find the diameter of the protractor.

Answer 1

Step1: Given information.

The perimeter of the semicircular protractor$=36\mathrm{cm}$

Step 2: Find the radius of the semicircular protractor.

The perimeter of the semicircle with the radius $r$ is,

$$\begin{gathered} \mathrm{Perimeter}=\mathrm{\pi }r+2r \\ \Rightarrow 36=\left(\frac{22}{7}\times r\right)+\left(2\times r\right) \\ \Rightarrow 36=r\times \left(\frac{22+\left(2\times 7\right)}{7}\right) \\ \Rightarrow 36=r\times \left(\frac{22+14}{7}\right) \\ \Rightarrow 36=r\times \left(\frac{36}{7}\right) \\ \Rightarrow \frac{36\times 7}{36}=r \\ \therefore r=7\mathrm{cm} \end{gathered}$$

Step 3: Find the diameter of the semicircular protractor.

Therefore, the diameter is,

$\begin{array}{rcl}d& =& 2r\\ & =& 2\times 7\\ & =& 14\mathrm{cm}\end{array}$

Thus, the diameter of the semicircular protractor is $14\mathrm{cm}$.