Question:
a wire when bent in the form of an equilateral triangle encloses an area of 121x1.73 sq.cm.If the same wire is bent in the form of a circle,find the area of the circle
Answer:
Let the side of the equilateral triangle = a
Noe area of the triangle = (√3/4)a2
Now (√3/4)a2 = 121*1.73
=> (1.73*a2 )/4 = 121*1.73 (since √3 = 1.73 )
=> a2 /4 = 121
=> a2 = 121*4
=> a = √(121*4)
=> a = 11*2
=> a = 22
Total length of the wire = 3a (since there are 3 sides in a traingle)
= 3*22
= 66
Now this wire is bent into circle.
Let radius of the circle is r.
So circumference of circle = 66
=> 2*Π*r = 66
=> (2*22*r)/7 = 66
=> 2*22*r = 66*7
=> r = (66*7)/(2*22)
=> r = (6*7)/(2*2)
=> r = (3*7)/2
=> r = 21/2
Now area of the circle = Π*r2
= (22/7)*(21/2)2
= (22*21*21)/(7*2*2)
= (11*21*21)/(7*2)
= (11*21*3)/2
= 693/2
= 346.5 cm2
So are of the circle is 346.5 cm2
Not what you are looking for? Go ahead and submit the question, we will get back to you.
