Q. 34.2( 352 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 (*r _{1}*) of gold region (i.e., 1st circle) = = 10.5 cm

Given:

Each circle is 10.5 cm wider than the previous circle.

Hence,

Radius (*r _{2}*) of 2nd circle = 10.5 + 10.5

21 cm

Radius (*r _{3}*) of 3rd circle = 21 + 10.5= 31.5 cm

Radius (*r _{4}*) of 4th circle = 31.5 + 10.5= 42 cm

Radius (*r _{5}*) of 5th circle = 42 + 10.5= 52.5 cm

Area of gold region = Area of 1^{st}circle

= πr1^{2}

= π (10.5)^{2}

= 346.5 cm^{2}

Area of red region = Area of 2nd circle - Area of 1^{st}circle

= πr_{2}^{2} – πr_{1}^{2}

= π (21)^{2} – π (10.5)^{2}

= 441π – 110.25π

= 330.75π

= 1039.5 cm^{2}

Area of blue region = Area of 3^{rd}circle - Area of 2^{nd}circle

= πr_{3}^{2} – πr2^{2}

= π (31.5)^{2} – π (21)^{2}

= 992.25 π - 441 π

= 551.25 π

= 1732.5 cm^{2}

Area of black region = Area of 4^{th}circle - Area of 3rd circle

= πr_{4}^{2} – πr_{3}^{2}

= π (42)^{2} – π (31.5)^{2}

= 1764 π – 992.25 π

= 771.75 π

= 2425.5 cm^{2}

Area of white region = Area of 5^{th}circle - Area of 4^{th}circle

= πr_{5}^{2} – πr_{4}^{2}

= π (52.5)^{2} – π (42)^{2}

= 2756.25 π - 1764 π

= 992.25 π

= 3118.5 cm^{2}

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm^{2}, 1039.5 cm^{2}, 1732.5 cm^{2}, 2425.5 cm^{2}, and 3118.5 cm^{2} respectively

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