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Three circles of radius 6 cm each touches each other externally. Then the distance of the centre of one circle from the line joining the centres of other two circles is equal to
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- 6√5 cm
- 6√3 cm
- 6√2 cm
- 6√7 cm
- 6√5 cm
Correct Option: B
According to question , we draw a figure of three circles of radius 6 cm each touches each other externally ,
ABC will be an equilateral triangle whose each side = 12 cm.
AD ⊥ BC
∴ BD = DC = 6 cm.
AD ⊥ BC
∴ AD = √AB² - BD²
AD= √12² - 6²
AD = √(12 + 6)(12 - 6)
AD = √18 × 6 = 6√3 cm.
