We read your photo as:

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

ScanSolve AI Answer

Answer

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

Step-by-step solution

6 steps

  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.

  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)

  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)

  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=3110 cmp = \frac{31}{10} \text{ cm}

  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=6625 cm2m^{2} = \frac{1225}{100} - \frac{961}{100} = \frac{264}{100} = \frac{66}{25} \text{ cm}^{2} m=6625=665 cmm = \sqrt{\frac{66}{25}} = \frac{\sqrt{66}}{5} \text{ cm}

  6. Length of common chord AB=2mAB = 2m

    AB=2×665=2665 cmAB = 2 \times \frac{\sqrt{66}}{5} = \frac{2\sqrt{66}}{5} \text{ cm} 2665 cm\dfrac{2\sqrt{66}}{5} \text{ cm}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QTwo circles of radii 3.5 and 4.2 cm with centres O1 and O2 intersect at points A and B as shown. The…QTwo circles centre's O1 and O2 are of the same radius, 8.75 cm. The two circles intersect one…QThe figure below shows two intersecting circles with radius 9 cm with centre O1 and O2. Find the…QTwo circles with centres O and Q and radii 8cm intersect at points A and B. The distance between O…QThe figure below shows two circles intersecting at A and D. The centres are A and C respectively. AB…
Browse by topic:Geometry
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.