Question

Fif. 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

SOLUTION:

GIVEN:

Diameter of Gold region= 21 cm

$\text{Radius of gold region=}$ 21/2= 10.5 cm

Area of Gold Region = πr²

= π(10.5)² =( 22/7)× 110.25 = 346.5 cm²

$\text{Area of Gold Region= 345.5}$ cm²

$\text{Radius of red region =}$ Radius for gold + red region= 10.5 + 10.5= 21 cm

Area of Red Region = π²(21² - 10.5²)

[Area of a ring= π (R²-r²), where R= radius of outer ring & r= radius of inner ring]

= 22/7 (21² – 10.5²) [ a²-b²= (a+b)(a-b)]

= 22/7 (21 + 10.5)(21 – 10.5)

= (22/7 )x 31.5 x 10.5 = 1039.5 cm²

$\text{Area of Red Region =}$ 1039.5 cm²

$\text{Radius of blue region =}$ Radius of blue region = Now radius for gold + red+ blue region= 21+10.5= 31.5 cm

Area of Blue Region = π(31.5² – 21²)

= 22/7 (31.5² - 21²)

= 22/7 (31.5 +21)(31.5 - 21)

= (22/7 )x 52.5 x 10.5 = 1732.5 cm²

$\text{Area of Blue Region =1732.5}$ cm²

Now,
$\text{Radius of black region}=$ radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

Area of Black Region = π(42² – 31.5²)

= 22/7 (42²-31.5² )

= 22/7 (42+31.5)(42-31.5)

= (22/7 )x 73.5 x 10.5 = 2425.5 cm²

$\text{Area of Black Region =2425.5}$ cm²

Now
$\text{Radius of white region}=$ radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

Area of White Region= π(52.5² – 42²)

= 22/7 (52.5²-42² )

= 22/7 (52.5+42)(52.5-42)

= (22/7 )x 94.5 x 10.5 = 3118.5 cm²

$\text{Area of white Region =3118.5}$ cm²