Question

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

Answer 1

Let $r_1$, $r_2$, $r_3$, $r_4$ and $r_5$ be the radii of the gold, red, blue, black and white regions, respectively.

Gold region

$d_1=2r_1=21cm$

$\therefore r_1=\frac{21}{2}cm$

$A_1=\pi {r_1}^2=\pi {\left(\frac{21}{2}\right)}^2$
$=\frac{441\pi }{4}c{m}^2$
$=\frac{441}{4}\times \frac{22}{7}c{m}^2$
$=\frac{63}{2}\times 11c{m}^2$
$=\frac{693}{2}c{m}^2$
$=346.5c{m}^2$

Hence, area of gold region $=346.5c{m}^2$

Red region

$r_2=r_1+10.5=\frac{21}{2}+\frac{21}{2}=21cm$

$A_2=\pi {r_2}^2=\pi (21{)}^2$
$=\frac{22}{7}\times 441c{m}^2$
$=22\times 63c{m}^2$
$=1386c{m}^2$

Hence, area of red region $=A_2-A_1=1039.5c{m}^2$

Blue region

$r_3=r_2+10.5=21+10.5=31.5cm$

$A_3=\pi {r_3}^2=\pi (31.5{)}^2$
$=\frac{22}{7}\times 992.25c{m}^2$
$=\frac{21829.5}{7}c{m}^2$
$=3118.5c{m}^2$

Area of blue region $=A_3-A_2=1732.5c{m}^2$

Black region

$r_4=r_3+10.5=31.5+10.5=42cm$

$A_4=\pi {r_4}^2=\pi (42{)}^2$
$=\frac{22}{7}\times 1764c{m}^2$
$=22\times 252c{m}^2$
$=5544c{m}^2$

Hence, area of black region $=A_4-A_3=2425.5c{m}^2$

White region

$r_5=r_4+10.5=42+10.5=42.5cm$

$A_5=\pi {r_5}^2=\pi (42.5{)}^2$
$=\frac{22}{7}\times 1806.25c{m}^2$
$=\frac{39737.5}{7}c{m}^2$
$=8662.5c{m}^2$

Hence, area of white region $=A_5-A_4=3118.5c{m}^2$