Properties Of A Chord Of A Circle I

Sum of the angles subtended
by the parallel chords at the centre = x + x/3 = 4x/3
um of the angles subtended
by them at any point on the remaining part of the circle = 4x/3 * 1/2 = 2x/3
(The angle subtended by an
arc at the centre is double the
angle at any point on the
circle)

Given, angle XOC = 41°
Angle BOC = 82°
Angle BOC = 2 * angle BAC
(The angle subtended by an
arc at the centre is double the
angle at any point on the
circle)
angle BAC = 82 /2 = 41

angle AOB = 104°
Angle ACB = Angle ADB = Angle AEB = angle AOB /2 = 104° /2 = 52°
(The angle subtended by an arc of the circle at the centre is double the angle subtended
by it at any point on the remaining part of the circle.)
x + y+ z = 52° + 52° + 52° = 156°

ABC is an equilateral triangle
angle BAC = 60
Angle BOC = 2 * angle BAC = 120
The angle subtended by an arc of the circle at the centre is double the angle subtended
by it at any point on the remaining part of the circle.

The angle subtended by an arc of the circle at the centre is double the angle subtended
by it at any point on the remaining part of the circle.
In case of a semicircle, Angle subtended at the centre is 180. Therefore, the angle subtended
by it at any point on the remaining part of the circle will be 180/2 = 90

Let AB be the chord of length 30 cm and C is the midpoint.
AC = 30 / 2 = 15 , OC = 8
In right triangle OAC, using Pythagoras theorem,
OA^2 = OC^2 + AC^2
OA^2 = 8^2 + 15^2
OA = 17
Let DE be the chord which is at a distance of 13 cm from the centre and F be the midpoint
In right triangle OFD, using Pythagoras theorem,
OD^2 = OF^2 + DF^2
17^2 = 13^2 + DF^2
DF = 2√30
DE = 2* 2√30 = 4√30

Let AB be the chord and C is the midpoint.
AC = 16 / 2 = 8 , OC = 9
In right triangle OAC, using Pythagoras theorem,
OA^2 = OC^2 + AC^2
OA^2 = 8^2 + 9^2
OA = √145

Let AB be the chord and C is the midpoint.
AC = 1/2 * AB = 1/2 * 10 = 5, OA = 13
In right triangle OAC, using Pythagoras theorem,
OA^2 = OC^2 + AC^2
13^2 = OC^2 + 5^2
OC = 12

As the length of the chord decreases then the distance of the chord from the center increases.
Smaller is the size of a chord, greater is its distance from the centre.

Let AB be the chord and C is the midpoint.
OC = 2, OA = 2√2
In right triangle OAC, using Pythagoras theorem,
OA^2 = OC^2 + AC^2
(2√2 )^2 = 2^2 + AC^2
AC = 2
AB = 2 * AC = 4