Properties Of A Chord Of A Circle – II

The consecutive angles of a parallelogram should be supplementary (180°).
Therefore, angle OQP + angle QPR = 180°
angle QPR = 180° – 110° = 70
Angle QPR = 2 * angle QTR
(The angle subtended by an
arc at the centre is double the
angle at any point on the
circle)
angle QTR = 70 /2 = 35
angle TQP = angle QTP = 20
(PQ and PT being the radii, Angles opposite the equal sides of a triangle are always equal.)
Similarly, angle PRT = x = angle PTR
angle QTR = angle QTP + angle PTR
35 = 20 + x
x = 15

Let the common chord be BC and O and A be the centers of the two circles.
BD = 1/2 BC = 1/2 * 5 = 2.5 (Perpendicular from center to the chord bisects the chord)
In right triangle, ABD
Using Pythagoras theorem,
AB^2 = AD^2 + BD^2
8^2 = AD^2 + 2.5^2
AD = 7.6
In right triangle, BDO
Using Pythagoras theorem,
OB^2 = OD^2 + BD^2
6^2 = OD^2 + 2.5^2
OD = 5.45
OA = OD + AD
OA = 13.05

Given, AB = 30, AE = 30/2 = 15
CD = 16, CF = 16/2 = 8
OE = 8
Let OF =x
In right triangle OEA,
AO^2 = AE^2 + OE^2
r^2 = 15^2 + 8^2
r=17
In right triangle OFC,
OC^2 = OF^2 + CF^2
17^2 = x^2 + 8^2
x=15

Given, arc PXQ : arc QYR = 8:5
∠POQ : ∠QOR = 8:5
∠POQ / ∠QOR = 8/5
128 : ∠QOR = 8/5
∠QOR = 128 * 5/8 = 80
Angle POR = 80+128 = 208

If the length of the chord is equal to the radius, the triangle which connects the two radii and the chord will be equilateral.
Therefore, any chord of length equal to its radius subtends
60 angle at the centre.
The central angle of a circle adds up to 360∘
360 / 60 = 6
This indicates that 6 equal chords can be drawn on a circle to form a regular hexagon inscribed in a circle.

Diameter ST is perpendicular to the chord and bisects the chord
Let O be the centre of the circle and r be the radius of the circle.
PQ = QR = PR/2 = 10/2 = 5
OQ = r – 3
In right triangle, OQP
Using Pythagoras theorem,
OP^2 = OQ^2 + PQ^2
r^2 = (r-3)^2 + 5^2
0 = -6r+9+25
r = 17/3

Let AB be the chord of the outer edge and C is the midpoint of AB.
To find the distance from C(inner edge) to B(outer edge)
OC = 21
OB = 29
In right triangle, OBC
Using Pythagoras theorem,
OB^2 = OC^2 + BC^2
29^2 = 21^2 + BC^2
BC = 20

Given, OA = 5
If an n-sided polygon in inscribed in a circle, then the angle subtented by each side of the polygon at the centre of the circle is 360/n
Therefore,
AOB = 90°
In right triangle, AOB
Using Pythagoras theorem,
AB^2 = AO^2 + OB^2
AB^2 = 2* AO^2
AB = 5√2
Perimeter of the swimming pool = 4 * 5√2= 20√2

Let R be any passenger seat on the wheel
Angle POQ = 2 * angle PQR
(The angle subtended by an
arc at the centre is double the
angle at any point on the
circle)
angle PQR = 126/2 = 63

Let AB be the chord and C is the midpoint and O be the center of the circle.
AC = 1/2 * AB = 1/2 * 6 = 3,
OA = 18/2 = 9
In right triangle OAC, using Pythagoras theorem,
OA^2 = OC^2 + AC^2
9^2 = OC^2 + 3^2
OC = 6√2