HCF of
20ab, 30bc
40 acb
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1
Radius (r1) of 1st circle = 19 cm
Radius (r2) or 2nd circle = 9 cm
Let the radius of 3rd circle be r.
Circumference of 1st circle = 2πr1 = 2π (19) = 38π
Circumference of 2nd circle = 2πr2 = 2π (9) = 18π
Circumference of 3rd circle = 2πr
Given that,
Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle
2πr = 38π + 18π = 56π
Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.
Answered by
2
Answer:
10b is the highest common factor of these no.
Hope this will help you
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