Question

Given figure depicts an archery target marked with its five scoring regions from the center outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is $ 21 cm$ and each of the other bands is $ 10.5 cm$ wide. Find the area of each of the five scoring regions.

CETEy L 

Answer 1

Step 1 : Finding the radius and area of gold region

Diameter of gold region$=21cm$

Therefore, radius$\left(r_1\right)$ of gold region$=\frac{21}{2}=10.5cm$

Assumption :$\mathrm{\pi }=3.14$

Area of gold region$=\mathrm{\pi }\times {{\mathrm{r}}_1}^2$

$=\mathrm{\pi }\times 10.5^2$

$\Rightarrow$ Area of gold region$=346.185c{m}^2$

Step 2 : Finding the radius and area of red region

Radius of red region $\left(r_2\right)=$radius of gold region $+$ width of the band

$$\begin{gathered} =10.5cm+10.5cm \\ =21cm \end{gathered}$$

Area of red region$=\mathrm{\pi }({{\mathrm{r}}_2}^2-{{\mathrm{r}}_1}^2)$ [Area of ring$=\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$; where $\mathrm{R}=$outer radius, $\mathrm{r}=$inner radius]

$=\mathrm{\pi }\left({21}^2-10.5^2\right)$

$=\mathrm{\pi }\left(21+10.5\right)\left(21-10.5\right)$ $\left[\right({\mathrm{a}}^2-{\mathrm{b}}^2)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)]$

$\Rightarrow$ Area of red region$=1038.555c{m}^2$

Step 3 : Finding the radius and area of blue region

Radius of blue region $\left(r_3\right)=$radius of red region $+$width of the band

$$\begin{gathered} =21cm+10.5cm \\ =31.5cm \end{gathered}$$

Area of blue region$=\mathrm{\pi }({{\mathrm{r}}_3}^2-{{\mathrm{r}}_2}^2)$ [Area of ring$=\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$; where $\mathrm{R}=$outer radius, $\mathrm{r}=$inner radius]

$=\mathrm{\pi }\left({31.5}^2-{21}^2\right)$

$=\mathrm{\pi }\left(31.5+21\right)\left(31.5-21\right)$ $\left[\right({\mathrm{a}}^2-{\mathrm{b}}^2)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)]$

$\Rightarrow$ Area of blue region$=1730.925c{m}^2$

Step 4 : Finding the radius and area of black region

Radius of black region $\left(r_4\right)=$radius of blue region $+$ width of the band

$$\begin{gathered} =31.5cm+10.5cm \\ =42cm \end{gathered}$$

Area of black region$=\mathrm{\pi }({{\mathrm{r}}_4}^2-{{\mathrm{r}}_3}^2)$ [Area of ring$=\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$; where $\mathrm{R}=$outer radius, $\mathrm{r}=$inner radius]

$\begin{array}{rcl}& =& \mathrm{\pi }({42}^2-31.5^2)\end{array}$

$\begin{array}{rcl}& =& \mathrm{\pi }(42+31.5)\left(42-31.5\right)\end{array}$ $\left[\right({\mathrm{a}}^2-{\mathrm{b}}^2)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)]$

$\Rightarrow$ Area of black region$=2423.295c{m}^2$

Step 5 : Finding the radius and area of white region

Radius of white region $\left(r_5\right)=$radius of black region $+$width of the band

$$\begin{gathered} =42cm+10.5cm \\ =52.5cm \end{gathered}$$

Area of white region$=\mathrm{\pi }({{\mathrm{r}}_5}^2-{{\mathrm{r}}_4}^2)$ [Area of ring$=\mathrm{\pi }\left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$; where $\mathrm{R}=$outer radius, $\mathrm{r}=$inner radius]

$=\mathrm{\pi }\left(52.5^2-{42}^2\right)$

$=\mathrm{\pi }\left(52.5+42\right)\left(52.5-42\right)$ $\left[\right({\mathrm{a}}^2-{\mathrm{b}}^2)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)]$

$\Rightarrow$ Area of white region$=3115.665c{m}^2$

Hence,

Area of gold region $=346.185c{m}^2$

Area of red region $=1038.555c{m}^2$

Area of blue region $=1730.925c{m}^2$

Area of black region$=2423.295c{m}^2$

Area of white region$=3115.665c{m}^2$