Math, asked by harshitktpyadav2004, 6 months ago

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
arcas
wide. Find the area of each of the five scoring regions.​

Answers

Answered by TheEternity
0

\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.

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