radius of a circle of is 10 cm. find the length of the chord if the chord is at a distance of 6 cm from the centre
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Answered by
4
Let the circle be with center O and radius 10 cm. Let there be a chord AB
Draw a perpendicular from O on AB to meet at P. The perpendicular from the center divides the chord in two halves. Given, OP=6cm
Thus, In △OAP, Using Pythagoras theorem
OA
2
=AP
2
+OP
2
10
2
=AP
2
+6
2
AP
2
=64
AP=8 cm
Thus, AB=2AP=16 cm
Now, new chord CD is drawn with length 8 cm. Draw a perpendicular on CD from O to cut CD at N.
Now, In △ONC
OC
2
=NC
2
+ON
2
10
2
=4
2
+ON
2
ON=
84
cm
Draw a perpendicular from O on AB to meet at P. The perpendicular from the center divides the chord in two halves. Given, OP=6cm
Thus, In △OAP, Using Pythagoras theorem
OA
2
=AP
2
+OP
2
10
2
=AP
2
+6
2
AP
2
=64
AP=8 cm
Thus, AB=2AP=16 cm
Now, new chord CD is drawn with length 8 cm. Draw a perpendicular on CD from O to cut CD at N.
Now, In △ONC
OC
2
=NC
2
+ON
2
10
2
=4
2
+ON
2
ON=
84
cm
Answered by
10
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Answer:
AB = 16 cm
Step-by-step explanation:
HERE:
OB is radius = 10
OC is the perpendicular distance from the chord to the centre of the circle = 6
AB is the given chord
TO FIND THE LENGTH OF THE CHORD
by pythagoras theorem
OC^2 + BC^2 = OB^2
6^2 + BC^2 = 10^2
BC^2 = 100 - 36 = 64
SO, BC = √64 = 8
We know that perpendicular bisector divides the chord into two equal parts
BC = AC = 8
BC + AC = 8 + 8
AB = 16 cm
HOPE IT IS USEFUL
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