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.
Answers
Question :-
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 :-
- area of gold region = 346.5 cm²
- area of red region = 1039.5 cm²
- area of blue region = 1732.5 cm²
- area of black region = 2425.5 cm²
- area of white region = 3118.5 cm²
Solution :-
Given -
- diameter of gold region = 21 cm
- width of each band = 10.5 cm
To Find -
area of each of the five scoring regions
Process -
We have,
Diameter of gold region = 21 cm
Therefore, radius of gold region = 21/2 cm
= 10.5 cm
Since width of each scoring region is 10.5 cm each, thus radii of different regions are as follows -
- radius of gold region = 1(10.5 cm)
- radius of red region = 2(10.5 cm)
- radius of blue region = 3(10.5 cm)
- radius of black region = 4(10.5 cm)
- radius of white region = 5(10.5 cm)
______________
Now,
area of gold region
= π(r of gold)²
= (22/7) × (10.5 cm)²
= [(22/7) × 10.5 × 10.5 ] cm²
= [22 × 1.5 × 10.5 ] cm²
= [22 × 15.75 ] cm²
= 346.5 cm²
______________
Again,
area of red region
= π [ (r of red)² - (r of gold)² ]
= π [ {2(10.5 cm)}² - {1(10.5 cm)}² ]
= π [ 4(10.5 cm)² - (10.5 cm)² ]
= π [ 3(10.5 cm)² ]
= [(22/7) × 3 × 10.5 × 10.5] cm²
= [346.5 × 3] cm²
= 1039.5 cm²
_______________
Again,
area of blue region
= π [ (r of blue)² - (r of red)² ]
= π [ {3(10.5 cm)}² - {2(10.5 cm)}² ]
= π [ 9(10.5 cm)² - 4(10.5 cm)² ]
= π [ 5(10.5 cm)² ]
= [(22/7) × 5 × 10.5 × 10.5] cm²
= [346.5 × 5] cm²
= 1732.5 cm²
_______________
Again,
area of black region
= π [ (r of black)² - (r of blue)² ]
= π [ {4(10.5 cm)}² - {3(10.5 cm)}² ]
= π [ 16(10.5 cm)² - 9(10.5 cm)² ]
= π [ 7(10.5 cm)² ]
= [(22/7) × 7 × 10.5 × 10.5] cm²
= [346.5 × 7] cm²
= 2425.5 cm²
_______________
At last,
area of white region
= π [ (r of white)² - (r of black)² ]
= π [ {5(10.5 cm)}² - {4(10.5 cm)}² ]
= π [ 25(10.5 cm)² - 16(10.5 cm)² ]
= π [ 9(10.5 cm)² ]
= [(22/7) × 9 × 10.5 × 10.5] cm²
= [346.5 × 9] cm²
= 3118.5 cm²
_______________
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