Question

Four students of class IX B with names Ajay. Babloo. Charan and Deepak are playing a
game in a circular playground.
All four students are holding radios with speaker and mic. These radios are connected by
a wire of equal length that is 11 m (for each radio). Ajay Asks a question to Babloo. Ir
Babloo gives the correct answer he gets 10 points and asks a new question to Charan, Ir
he can not answer then he passes the same question to Charan and gets no points.
These conditions apply to all four players. After 10 rounds who gets maximum points, he
becomes the winner
Ay
Deepak
Charan
| What is the radius of the field?
a 7 m
b. 14 m
11 m
d. 22 m
IL What is the area of the field?
a 70 m
b. 154 m
c 110 m
d. 220 m
iii. What is the area of the part marked with 1 on the field?
a. 50 m
b. 154 m
C 76m
d. 38.5 m
iv. What is the circumference of the field?
a. 22 m
b. 14 m

Answer 1

Answer:

1. 7m
2. 154 msq
3. 38.5 m sq
4. 44 m
5. 14 m

Answer 2

Given : Four students of class IX B with names Ajay, Babloo, Charan and I

in a circular playground.  

All four students are holding radios with speaker and mic. These

radios are connected by a wire of equal length that is 11 m (for each radio)

To Find :

What is the radius of the field?

a 7 m

b. 14 m

11 m

d. 22 m

What is the area of the field?

a 70 m²

b. 154 m²

c 110 m²

d. 220 m²

What is the area of the part marked with 1 on the field?

a. 50 m²

b. 154 m²

C 76m²

d. 38.5 m²

iv. What is the circumference of the field?

a. 22 m

b. 14 m

c 44 m

d  28 m

Direct distance from ajay to charan

a 7 m

b  28 m

c 15 m

d  14 m

Solution:

radius of the field  = R

Then  (90/360) * 2πR  = 11

=> 2πR = 44

=> 2(22/7) R = 44

=> R = 7

radius of the field = 7m

Area of field = πR² = (22/7)7² = 154 m²

area of the part marked with 1 on the field = (90/360) * 154

= 154/4

= 77/2

= 38.5 m²

circumference of the field = 2πR  = 2(22/7) * 7 = 44 m

Direct distance from ajay to charan = Diameter = 2R = 2 * 7 = 14 m

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