Question : To keep the lawn green and cool, Sadhna uses water sprinklers which rotate in circular shape and cover a particular area. The diagram below shows the circular areas covered by two sprinklers:
Two circles touch externally. The sum of their areas is 130π sq. m and the distance between their centres is 14 m.
Based on the above information, answer the following questions :
(i) Obtain a quadratic equation involving R and r from above. [1 Mark]
(ii) Write a quadratic equation involving only r. [1 Mark]
(iii) (a) Find the radius r and the corresponding area irrigated. [2 Marks]
OR
(b) Find the radius R and the corresponding area irrigated. [2 Marks]
Solution :
(i) Circular Area covered by First Sprinkler + Circular Area covered by Second Sprinkler = 130π m² (Given)
πR²+πr²=130π
π(R²+r²)=130π
R²+r²=130 — (1)
which is the required quadratic equation involving R and r.
(ii) R+r=14 m (Given)
R=14-r — (2)
Substituting in equation (1)
R²+r²=130
(14-r)²+r²=130
(14)²+(r)²-2(14)(r)+r²=130
196+r²-28r+r²=130
2r²-28r+196=130
2r²-28r+196-130=0
2r²-28r+66=0
2[r²-14r+33]=0
r²-14r+33=0/2
r²-14r+33=0
which is the required quadratic equation involving only r.
(iii)
(a) r²-14r+33=0
Solving by Factorisation Method (Splitting the Middle Term)
r²-(11+3)r+33=0
r²-11r-3r+33=0
r(r-11)-3(r-11)=0
(r-11)(r-3)=0
r-11=0
r=11
but we know R+r=14 m and r is smaller radius so r≠11
r-3=0
r=3 m
Hence, the required value of r is 3 m
Corresponding Area irrigated
= πr²
= π(3)²
=9π m²
(b) We know,
r = 3 m
Also
R+r=14
R+3=14
R=14-3
R=11 m
Corresponding Area Irrigated
=πR²
=π(11)²
=π(121)
=121π m²