AREAS RELATED TO CIRCLES(solution) Ncert & Cbse
EXERCISE 12.1
Q.1 The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
solution:- Given Radii of Two Circles
R1 =19 cm and R2 =9 cm
According to Question,
the sum of the circumferences of the two circles=2πr
⇒2πR1 + 2πR2=2πr
⇒2π(R1 + R2)=2πr
⇒2π(19 cm+ 9 cm)=2πr
⇒r=28cm
Radius of circle=28cm
Q.2 The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
solution:- Given Radii of Two Circles
R1 = 8 cm and R2 = 6 cm
According to Question,
the sum of the Area of the two circles=πr^2
⇒π(R1)^2 + π(R2)^2=πr^2
⇒π{(8)^2 + (6)^2}= π r^2
⇒64 + 36=r^2
⇒100= r^2
⇒r=√100
⇒r=10 cm
The radius of the circle= 10 cm ans
Q.3 Fig. 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.
Soln:- Given that Diameter of gold region=21 cm
then radius of Gold region=21/2 cm=10.5 cm
and also given that wide of bands of each regions =10.5 cm
know area of gold region=πr^2
=(22/7) (10.5)^2=(2,425.5)/7 cm^2 =346.5 cm^2
area of Red region=π(10.5+10.5)^2 - area of gold region
= π(21)^2- 346.5
=(441)(22/7) -346.5
=1,386-346.5 =1,039.5 cm^2
area of Blue region=π(10.5+10.5+10.5)^2-(area of Red region+area of gold region)
= π(31.5)^2-(346.5+1,039.5)
= (22/7)(31.5)^2-1,386
=3,118.5- 1,386=1,732.5 cm^2
area of Black region=π(10.5+10.5+10.5+10.5)^2-(area of gold region+area of Red region + area of Blue region)
area of Black region=π(10.5+10.5+10.5+10.5)^2-(346.5+1,039.5+1,732.5)
= (22/7)(42)^2-3,118.5
=5,544 -3,118.5
=2,425.5 cm^2
area of white region=π(10.5+10.5+10.5+10.5+10.5)^2-(area of gold region+area of Red region + area of Blue region+area of Black region)
area of white region=π(52.5)^2-(346.5+1,039.5+1,732.5+2,425.5)
=(22/7)( 52.5)^2-8,662.5
=8,662.5 -5,544
=3,318.5
hence area of the each regions
Gold : 346.5 cm^2
Red : 1039.5 cm^2
Blue : 1732.5 cm^2
Black : 2425.5 cm^2
White : 3118.5 cm^2
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