Question

3. Fig. 12.3 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

$\large{\underline{\rm{\green{\bf{Given:-}}}}}$

The diameter of the region representing  gold score = 21 cm

Width of each other bands = 10.5 cm

$\large{\underline{\rm{\green{\bf{To \: Find:-}}}}}$

Find the area of each of the five scoring regions.

$\large{\underline{\rm{\green{\bf{Solution:-}}}}}$

The radius of 1st circle, $\sf r_1 = \dfrac{21}{2} \: cm$ (as diameter D is given as 21 cm)

So, area of gold region = $\sf \pi {r_1}^{2} = \pi (10.5)^{2}$

Substituting their values,

Area of gold region = $\sf \pi (10.5)^{2}=346.5 \: cm^{2}$

Now, it is given that each of the other bands is 10.5 cm wide,

So, the radius of 2nd circle, $\sf r_2$

$\implies \sf 10.5 \: cm+10.5\: cm = 21\: cm$

Thus,

∴ Area of red region = Area of 2nd circle − Area of gold region = $\sf (\pi {r_2}^{2}-346.5) \: cm^{2}$

$\implies \sf = (\pi (21)^{2} - 346.5)\: cm^{2}$

$\implies \sf 1386 - 346.5$

$\implies \sf 1039.5\: cm^{2}$

Similarly,

The radius of 3rd circle, $\sf r_3$ = 21 cm + 10.5 cm = 31.5 cm

The radius of 4th circle, $\sf r_4$ = 31.5 cm + 10.5 cm = 42 cm

The Radius of 5th circle, $\sf r_5$ = 42 cm + 10.5 cm = 52.5 cm

For the area of nth region,

A = Area of circle n – Area of circle (n-1)

∴ Area of blue region (n=3) = Area of third circle – Area of second circle

$\implies \sf \pi (31.5)^{2} - 1386 \: cm^{2}$

$\implies \sf 3118.5 - 1386 \: cm^{2}$

$\implies \sf 1732.5 \: cm^{2}$

Therefore, the area of blue region is 1732.5 cm²

∴ Area of black region (n=4) = Area of fourth circle – Area of third circle

$\implies \sf \pi (42)^{2} - 1386\: cm^{2}$

$\implies \sf 5544 - 3118.5\: cm^{2}$

$\implies \sf 2425.5 \: cm^{2}$

Therefore, the area of the black region is 2425.5 cm²

∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle

$\implies \sf \pi (52.5)^{2} - 5544\: cm^{2}$

$\implies \sf 8662.5 - 5544\: cm^{2}$

$\implies \sf 3118.5 \: cm^{2}$

Therefore the area of white region is 3118.5 cm²

Answer 2

$\huge\color{red}{\underline{\underline {Given}}}$

ᴅɪᴀᴍᴇᴛᴇʀ ᴏꜰ ɢᴏʟᴅ ʀᴇɢɪᴏɴ= 21 ᴄᴍ

ᴛʜᴇɴ, ʀᴀᴅɪᴜꜱ ᴏꜰ ɢᴏʟᴅ ʀᴇɢɪᴏɴ =21/2 = 10.5 ᴄᴍ

$\huge\color{red}{\underline{\underline {To \: Find}}}$

The area of each of the five scoring regions.

$\huge\color{red}{\underline{\underline {Solution}}}$

ᴀʀᴇᴀ ᴏꜰ ɢᴏʟᴅ ʀᴇɢɪᴏɴ

= πʀ^2 = 22/7 (10.5)^2

= 22/7 x 110.25

= 346.5 ᴄᴍ ꜱq.

ʀᴀᴅɪᴜꜱ ꜰᴏʀ ɢᴏʟᴅ + ʀᴇᴅ ʀᴇɢɪᴏɴ

= 10.5 + 10.5

= 21 ᴄᴍ [ ʙʏ ɢɪᴠᴇɴ ᴄᴏɴᴅɪᴛɪᴏɴ ]

•°• ᴀʀᴇᴀ ꜰᴏʀ ʀᴇᴅ ʀᴇɢɪᴏɴ

= 22/7 [(21)^2 - (10.5)^2]

[ °•° ᴀʀᴇᴀ ᴏꜰ ᴀ ʀɪɴɢ = π (ʀ^2 - ʀ^2)

ᴡʜᴇʀᴇ, ʀ = ʀᴀᴅɪᴜꜱ ᴏꜰ ᴏᴜᴛᴇʀ ʀɪɴɢ, ᴀɴᴅ

ʀ = ʀᴀᴅɪᴜꜱ ᴏꜰ ɪɴɴᴇʀ ʀɪɴɢ.]

= 22/7 (21)^2 - 22/7 (10.5)^2

= 1386 - 346.5

= 1039.5 ᴄᴍ ꜱq.

ɴᴏᴡ, ʀᴀᴅɪᴜꜱ ꜰᴏʀ ɢᴏʟᴅ + ʀᴇᴅ + ʙʟᴜᴇ ʀᴇɢɪᴏɴ = 21 + 10.5 = 31.5 ᴄᴍ [ ʙʏ ɢɪᴠᴇɴ ᴄᴏɴᴅɪᴛɪᴏɴ ]

•°• ᴀʀᴇᴀ ꜰᴏʀ ʙʟᴜᴇ ʀᴇɢɪᴏɴ

= 22/7 [(31.5)^2 - (21)^2]

= 22/7 ( 31.5)^2 - 22/7(21)^2

= 3118.5 - 1386

= 1732.5 ᴄᴍ ꜱq.

ɴᴏᴡ, ʀᴀᴅɪᴜꜱ ꜰᴏʀ ɢᴏʟᴅ + ʀᴇᴅ + ʙʟᴜᴇ + ʙʟᴀᴄᴋ ʀᴇɢɪᴏɴ

= 31.5 + 10.5

= 42 ᴄᴍ [ ʙʏ ɢɪᴠᴇɴ ᴄᴏɴᴅɪᴛɪᴏɴ ]

•°• ᴀʀᴇᴀ ꜰᴏʀ ʙʟᴀᴄᴋ ʀᴇɢɪᴏɴ

= 22/7 [(42)^2 - (31.5)^2]

= 22/7 (42)^2 - 22/7 (31.5)^2

= 5544 - 3118.5

= 2425.5 ᴄᴍ ꜱq.

ʀᴀᴅɪᴜꜱ ꜰᴏʀ ɢᴏʟᴅ + ʀᴇᴅ + ʙʟᴜᴇ + ʙʟᴀᴄᴋ + ᴡʜɪᴛᴇ ʀᴇɢɪᴏɴ

= 42 + 10.5

= 52.5 ᴄᴍ [ ʙʏ ɢɪᴠᴇɴ ᴄᴏɴᴅɪᴛɪᴏɴ ]

•°• ᴀʀᴇᴀ ꜰᴏʀ ᴡʜɪᴛᴇ ʀᴇɢɪᴏɴ

= 22/7 [(52.5)^2 - (42)^2]

= 22/7 (52.5)^2 - 22/7 (42)^2

= 8662.5 - 5544

= 3118.5 ᴄᴍ ꜱq.

$\huge\color{red}{\underline{\underline {Direct \:Answer}}}$

GOLD - 346.5 cm sq.

RED - 1039.5 cm sq.

BLUE - 1732.5 cm sq.

BLACK - 2425.5 cm sq.

WHITE - 3118.5 cm sq.