## Areas related to Circles ncert solutions Chapter 12 Exercise 12.1 Question 3

Areas related to Circles ncert solutions Chapter 12 Exercise 12.1 Question 3

3.    Fig depicts an archery target marked with its five scoring regions from the center 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.

Solution:

Diameter of gold circle = 21 cm

Therefore, radius of gold circle (r1) = $\frac{21}{2}=10.5$ cm

Area of Gold circle = $\pi (r1)^2=\frac{22}{7} \times (10.5)^2= 346.5$ $cm^2$

Area of Red portion = Area of second circle from inside - area of gold circle =$\pi (21)^2-\pi (10.5)^2=1386-346.5=1039.5$ $cm^2$

Radius of third circle from inside = radius of gold circle + 10.5 +10.5

=10.5+10.5+10.5=31.5 cm

Area of Blue portion = Area of third circle from inside - area of second circle from inside=$\pi (31.5)^2-1386=3118.5-1386=1732.5$ $cm^2$

Radius of fourth circle from inside = radius of golden circle +10.5+10.5+10.5

=10.5+10.5+10.5+10.5=42 cm

Area of Black portion = Area of fourth circle from inside - area of third circle from inside=$\pi (42)^2-3118.5=5544-3118.5=2425.5$ $cm^2$

=10.5+10.5+10.5+10.5+10.5=52.5 cm

Area of white portion = Area of outer circle - area of fourth circle from inside =$\pi (52.5)^2 -5544=8662.5-5544=3118.5$ $cm^2$

### One Response to “Areas related to Circles ncert solutions Chapter 12 Exercise 12.1 Question 3”

• harsh says:

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