Question

Question 3
Figure 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 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Answer 1

Diameter of Gold circle (first circle) = 21 cm
Radius of first circle, $r_1=\frac{21}{2}cm=10.5cm$
Each of the other bands is 10.5 cm wide,
$\therefore$ Radius of second circle, $r_2=10.5cm+10.5cm=21cm$
$\therefore$ Radius of third circle, $r_3=21cm+10.5cm=31.5cm$
$\therefore$ Radius of fourth circle, $r_4=31.5cm+10.5cm=42cm$
$\therefore$ Radius of fifth circle, $r_5=42cm+10.5cm=52.5cm$
Area of gold region $=\pi r_1^2=\pi (10.5{)}^2=346.5c{m}^2$
Area of red region = Area of second circle - Area of first circle
= $\pi r_2^2-346.5c{m}^2=\pi (21{)}^2-346.5c{m}^2$
= $1386-346.5c{m}^2=1039.5c{m}^2$
Area of blue region = Area of third circle - Area of second circle
$=\pi r_3^2-1386c{m}^2=\pi (31.5{)}^2-1386c{m}^2$
$=3118.5-1386c{m}^2=1732.5c{m}^2$
Area of black region = Area of fourth circle - Area of third circle
$=\pi r_3^2-3118.5c{m}^2=\pi (42{)}^2-1386c{m}^2$
$=5544-3118.5c{m}^2=2425.5c{m}^2$
Area of white region = Area of fifth circle - Area of fourth circle
= $\pi r_4^2-5544c{m}^2=\pi \left(52.5\right)2-5544c{m}^2$
= $8662.5-5544c{m}^2=3118.5c{m}^2$