Two circles of radii 3.5 and 4.2 cm with centres O1 and O2 respectively intersect at points A and B as shown in the figure below. The distance between the two centres is 6 cm.

Mathematics
Two circles of radii 3.5 and 4.2 cm with centres O1 and O2 respectively intersect at points A and B as shown in the figure below. The distance between the two centres is 6 cm.

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Answer

2665 cm\dfrac{2\sqrt{66}}{5} \text{ cm}

Step 1: Let the centres be OO and QQ with OQ=5OQ = 5 cm, radius OA=3.5OA = 3.5 cm, and radius QB=2.5QB = 2.5 cm, where AA and BB are the points of intersection.
Let MM be the midpoint of common chord ABAB, so AM=MB=mAM = MB = m cm.
The line OQOQ is perpendicular to ABAB at MM. Let OM=pOM = p cm, so QM=5−pQM = 5 - p cm.

Step 2: Apply Pythagoras theorem in △OMA\triangle OMA:

OA2=OM2+AM2OA^{2} = OM^{2} + AM^{2} (3.5)2=p2+m2\left(3.5\right)^{2} = p^{2} + m^{2} (72)2=p2+m2\left(\frac{7}{2}\right)^{2} = p^{2} + m^{2} 494=p2+m2(1)\frac{49}{4} = p^{2} + m^{2} \quad (1)

Step 3: Apply Pythagoras theorem in △QMB\triangle QMB:

QB2=QM2+MB2QB^{2} = QM^{2} + MB^{2} (2.5)2=(5−p)2+m2\left(2.5\right)^{2} = \left(5 - p\right)^{2} + m^{2} (52)2=(5−p)2+m2\left(\frac{5}{2}\right)^{2} = \left(5 - p\right)^{2} + m^{2} 254=(5−p)2+m2(2)\frac{25}{4} = \left(5 - p\right)^{2} + m^{2} \quad (2)

Step 4: Subtract equation (2) from equation (1):

494−254=p2+m2−[(5−p)2+m2]\frac{49}{4} - \frac{25}{4} = p^{2} + m^{2} - \left[ \left(5 - p\right)^{2} + m^{2} \right] 244=p2−(5−p)2\frac{24}{4} = p^{2} - \left(5 - p\right)^{2} 6=p2−(25−10p+p2)6 = p^{2} - \left(25 - 10p + p^{2}\right) 6=p2−25+10p−p26 = p^{2} - 25 + 10p - p^{2} 6=10p−256 = 10p - 25 10p=3110p = 31 p=3110cmp = \frac{31}{10} cm

Step 5: Substitute p=3110p = \frac{31}{10} into equation (1):

m2=494−(3110)2m^{2} = \frac{49}{4} - \left(\frac{31}{10}\right)^{2} m2=494−961100m^{2} = \frac{49}{4} - \frac{961}{100} m2=1225100−961100=264100=6625cm2m^{2} = \frac{1225}{100} - \frac{961}{100} = \frac{264}{100} = \frac{66}{25} cm^{2} m=6625=665cmm = \sqrt{\frac{66}{25}} = \frac{\sqrt{66}}{5} cm

Step 6: Length of common chord AB=2mAB = 2m:

AB=2×665=2665cmAB = 2 \times \frac{\sqrt{66}}{5} = \frac{2\sqrt{66}}{5} cm

\dfrac{2\sqrt{66}{5} cm}

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Let the centres be O and Q with OQ = 5 cm, radius OA = 3.5 cm, and radius QB = 2.5 cm, where A and B are the points of intersection.

Two circles of radii 3.5 and 4.2 cm with centres O1 and O2 respectively intersect at points A and B as shown in the figure below. The distance between the two centres is 6 cm.
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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Step 1: Let the centres be O and Q with OQ = 5 cm, radius OA = 3.5 cm, and radius QB = 2.5 cm, where A and B are the points of intersection. Let M be the midpoint of common chord AB, so AM = MB = m cm. The line OQ is perpendicular to AB at M. Let OM = p cm, so QM = 5 - p cm. Step 2: Apply Pythagoras theorem in OMA: OA^2 = OM^2 + AM^2 (3.5)^2 = p^2 + m^2 ((7)/(2))^2 = p^2 + m^2 (49)/(4) = p^2 + m^2 (1) Step 3: Apply Pythagoras theorem in QMB: QB^2 = QM^2 + MB^2 (2.5)^2 = (5 - p)^2 + m^2 ((5)/(2))^2 = (5 - p)^2 + m^2 (25)/(4) = (5 - p)^2 + m^2 (2) Step 4: Subtract equation (2) from equation (1): (49)/(4) - (25)/(4) = p^2 + m^2 - [ (5 - p)^2 + m^2 ] (24)/(4) = p^2 - (5 - p)^2 6 = p^2 - (25 - 10p + p^2) 6 = p^2 - 25 + 10p - p^2 6 = 10p - 25 10p = 31 p = (31)/(10) cm Step 5: Substitute p = (31)/(10) into equation (1): m^2 = (49)/(4) - ((31)/(10))^2 m^2 = (49)/(4) - (961)/(100) m^2 = (1225)/(100) - (961)/(100) = (264)/(100) = (66)/(25) cm^2 m = sqrt((66)/(25)) = sqrt(66)5 cm Step 6: Length of common chord AB = 2m: AB = 2 × sqrt(66)5 = 2sqrt(66)5 cm 2sqrt(66)5 cm