The area of circle whose radius is 8cm is trisected by two concentric circles. The ratio of radii of the concentric circles in a scending order isa)1 : 2 : 3b)2 : 3 : 5c)1 :√2 :√3d)√2 :√3 :√5Correct answer is option 'C'. Can you explain this answer?

Question

The area of circle whose radius is 8cm is trisected by two concentric circles. The ratio of radii of the concentric circles in a scending order isa)1 : 2 : 3b)2 : 3 : 5c)1 :&radic;2 :&radic;3d)&radic;2 :&radic;3 :&radic;5Correct answer is option 'C'. Can you explain this answer?

Answer

The area of the original circle is given by A = πr^2, where r = 8cm. Therefore, the area of the original circle is A = π(8^2) = 64π cm^2.

If two concentric circles trisect the area of the original circle, then each of the three sections has an area of 64π/3 cm^2.

Let the radius of the innermost circle be r1, the radius of the middle circle be r2, and the radius of the outermost circle be r3. Then we have:

Area of innermost section = πr1^2 = 64π/3
Area of middle section = π(r2^2 - r1^2) = 64π/3
Area of outermost section = π(r3^2 - r2^2) = 64π/3

Simplifying each of these equations, we get:

r1^2 = 64/3
r2^2 - r1^2 = 64/3
r3^2 - r2^2 = 64/3

Adding the first two equations, we get:

r2^2 = 128/3

Adding the second two equations, we get:

r3^2 = 192/3 = 64

Therefore, the ratio of the radii of the concentric circles in ascending order is:

r1 : r2 : r3 = √(64/3) : √(128/3) : √64
= 4/√3 : 8/√3 : 8
= (4/√3)(√3/√3) : (8/√3)(√3/√3) : (8)(√3/√3)
= 4√3 : 8√3 : 8√3
= 4 : 8 : 8

Simplifying this ratio by dividing each term by 4, we get:

1 : 2 : 2

Therefore, the correct answer is (C) 1 : 2 : 2.

Answer

C

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