Q. 34.1( 395 Votes )

# Fig. 12.3 depicts an archery target marked with its five scoring areas 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 :

Radius (r1) of gold region (i.e., 1st circle) = = 10.5 cm

Given:

Each circle is 10.5 cm wider than the previous circle.

Hence,

Radius (r2) of 2nd circle = 10.5 + 10.5

21 cm

Radius (r3) of 3rd circle = 21 + 10.5= 31.5 cm

Radius (r4) of 4th circle = 31.5 + 10.5= 42 cm

Radius (r5) of 5th circle = 42 + 10.5= 52.5 cm

Area of gold region = Area of 1stcircle

= πr12

= π (10.5)2

= 346.5 cm2

Area of red region = Area of 2nd circle - Area of 1stcircle

= πr22 – πr12

= π (21)2 – π (10.5)2

= 441π – 110.25π

= 330.75π

= 1039.5 cm2

Area of blue region = Area of 3rdcircle - Area of 2ndcircle

= πr32 – πr22

= π (31.5)2 – π (21)2

= 992.25 π - 441 π

= 551.25 π

= 1732.5 cm2

Area of black region = Area of 4thcircle - Area of 3rd circle

= πr42 – πr32

= π (42)2 – π (31.5)2

= 1764 π – 992.25 π

= 771.75 π

= 2425.5 cm2

Area of white region = Area of 5thcircle - Area of 4thcircle

= πr52 – πr42

= π (52.5)2 – π (42)2

= 2756.25 π - 1764 π

= 992.25 π

= 3118.5 cm2

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm2, 1039.5 cm2, 1732.5 cm2, 2425.5 cm2, and 3118.5 cm2 respectively

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