Question

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

Answer 1

width of each circle $l=150cm$

circle $A$:- diameter $=k=21cm$

circle $B\Rightarrow$ diametre $=k+2l=42cm=2k$

circle $C\Rightarrow$ diametre $=k+4l=21\times 3cm=3k$

circle $D\Rightarrow$ diametre $=k+6l=21\times 4cm=4k$

circle $3$ dimeter $=k+8l=21\times 5cm=5k$

area of gold region = area of circle $A=\frac{\pi k^2}{4}$

$=\frac{22}{4}\times \frac{21\times 21}{4}=346.5c{m}^2$

are of red region = area of circle $B-$ area of circle $A$

$=\frac{\pi (2k{)}^2}{4}-\frac{\pi (k{)}^2}{4}=3x\frac{\pi k^2}{4}=3346.5c{m}^2$

$=1039.5c{m}^2$

area of blue region $=areaofcircle$C-$areofcircle$B$

$=\frac{\pi (3k{)}^2}{4}-\frac{\pi (2k{)}^2}{4}=5\times \left(\frac{\pi k^2}{4}\right)=5\times 346.5c{m}^2$

$=1732.5c{m}^2$

area of black region = are of circle $D-$ area of circle $C$

$=\frac{\pi (4k{)}^2}{4}-\frac{\pi (3k{)}^2}{4}=3dfrac\pi k^{2}4=7\times 346.5c{m}^2$

$=2425.5c{m}^2$

area of white region = area of circle $E-$ area of circle $D$

$=\frac{\pi (5k{)}^2}{4}-\frac{\pi (4k{)}^2}{4}=9\frac{\pi k^2}{4}$

$=9\times 346.5c{m}^2=3118.5c{m}^2$

$\therefore$ The area $346.5,1039.5,1732.5,2425.5$ and $3118.5c{m}^2$ respectively