Question

Three carom board strikers of radius 3.5 cm are so arranged such that each strikers, then the area of the empty space between the strikers is.
a) 10.5 m2
b) 38.5 m2
c) 1.967 cm2 d) 19.5 cm2

Answer 1

Answer:

Step-by-step explanation:

Given, A,B and C  are the centres of three circles touching each other at P,Q& R. All the circles have equal radii r=3.5 cm.

To find out: The area enclosed by the circles.

Solution:

We join AB,BC and  CA. Since AB joins the centres of two circles touching each other, we have AB= sum of their radii =2r=2×3.5 cm =7 cm.

Similarly BC=7 cm and CA=7 cm.

∴ΔABC is equilateral with sides =a=7 cm.  

So area ΔABC=  

4

3

​

 

​

×a  

2

=  

4

3

​

 

​

×7  

2

cm  

2

=  

4

49  

3

​

 

​

×cm  

2

=21.217 cm  

2

.  

Also each angle of ΔABC=60  

o

.

Now AQ&AP are the radii of the same circle of which

PQ is an arc.

∴APQ is a sector of central angle =θ=60  

o

 and radius =r=3.5 cm.

∴ Area of sector APQ=  

360  

o

 

θ

​

×π×r  

2

=  

360  

o

 

60  

o

 

​

×  

7

22

​

×3.5  

2

cm  

2

=6.417 cm  

2

.

Since all the sectors have same radii and central angle they are equal in areas.

∴ Area of three sectors =3×6.417 cm  

2

=19.251 cm  

2

.

∴ Area of enclosed region = area .ΔABC− Area of three sectors =(21.217−19.251)cm  

2

=1.966 cm  

2