in a circle , a chord 3 centimeters away from the center is 8 centimeters long . how long is a chord ,1.4 centimeters away from the center in the circle ?
Answers
Answered by
0
Let chord 8cm long meet the circle at at points A , B. Let Perpendicular from centre,say O, meet the above chord at C. Point C divides AB in two equal parts at C,i.e., AC = BC.
Join OA. Then Triangle ACO have sides 4cm, 3cm and radius r. Then Radius = √(4² + 3²) = 5.
Now draw a chord 1.4cm from centre O. Let this chord meet circle at points D, E.
Let perpendicular 1.4cm long meet chord DE at F. Join OD which is = radius = 5.
Now in triangle ODF, OD = 5, OF = 1.4 . Hence DF = √(OD² - OF²) = √(5² - 1.4²) = 4.8. Hence length of the chord = 2x4.8 = 9.6cm
Complete triangle
Join OA. Then Triangle ACO have sides 4cm, 3cm and radius r. Then Radius = √(4² + 3²) = 5.
Now draw a chord 1.4cm from centre O. Let this chord meet circle at points D, E.
Let perpendicular 1.4cm long meet chord DE at F. Join OD which is = radius = 5.
Now in triangle ODF, OD = 5, OF = 1.4 . Hence DF = √(OD² - OF²) = √(5² - 1.4²) = 4.8. Hence length of the chord = 2x4.8 = 9.6cm
Complete triangle
Answered by
0
In the figure, OM = 3 cm
AB = 8 cm
Since OM bisects AB
Therefore, AM = BM = 4 cm
In the right triangle OMA
OM = 3 cm, AM = 4 cm
Therefore, by Pythagorean theorem, OA = 5 cm
That is, the radius of the circle = 5 cm
Now, ON = 1.4 cm
Let CN = x cm
By pythagorean theorem
x = 4.8
Therefore, the length of the chord CD = 2x = 9.6 cm
AB = 8 cm
Since OM bisects AB
Therefore, AM = BM = 4 cm
In the right triangle OMA
OM = 3 cm, AM = 4 cm
Therefore, by Pythagorean theorem, OA = 5 cm
That is, the radius of the circle = 5 cm
Now, ON = 1.4 cm
Let CN = x cm
By pythagorean theorem
x = 4.8
Therefore, the length of the chord CD = 2x = 9.6 cm
Attachments:
Mathexpert:
Hope you like the solution. Please rate it 5 star if it is helpful.
Similar questions
Science,
8 months ago
Hindi,
1 year ago
Social Sciences,
1 year ago
English,
1 year ago
Chemistry,
1 year ago