Question

The art department of a school prepared circular hanging for the annual function of
the school. The circle with diameter 42 cm was placed at the bottom and the size of the middle circle was half the size of the bottom one and the smallest circle was half the size of the middle circle.​
(i)the radius of the small circle is?
(ii)the area of middle circle is?
(iii) the area of bottom circle region that is visible is?
(iv) the ratio of areas of three circles, starting from the bottom circle is?
(v)a colourful tape was used to decorate the boundary of the bottom circle.the length of tape used is?

Answer 1

Given : The art department of a school prepared circular hanging for the annual function of the school.

The circle with diameter 42 cm was placed at the bottom and the size of the middle circle was half the size of the bottom one and the smallest circle was half the size of the middle circle.​

To Find : (i)the radius of the small circle is?

(ii)the area of middle circle is?

(iii) the area of bottom circle region that is visible is?

(iv) the ratio of areas of three circles, starting from the bottom circle is?

(v) a colorful tape was used to decorate the boundary of the bottom circle.  the length of tape used is

Solution:

Bottom circle Diameter = 42 cm

=> Radius = 42/2  = 21 cm

Size = Area = πr²  = (22/7) (21)²  = 1386 cm²  

Area of middle circle = (1/2) 1386  =  693 cm²

Area of small circle = 693/2

(22/7)r² = 693/2

=>r² =  441/4

=> r = 21/2   cm

the area of bottom circle region that is visible is = area of larger - area of middle

=  1386 - 693

= 693   cm²

the ratio of areas of three circles, starting from the bottom circle is

1386 : 693 : 693/2

2 :  1 :  1/2

4  : 2  :  1

a colorful tape was used to decorate the boundary of the bottom circle.  the length of tape used is   = 2πr

= 2 * 22/7  * 21

= 44 * 3

= 132 cm

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