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Circle Theorem
Circle theorem rules:
1. The angle which an arc of a circle subtends at the centre of a circle is
twice the angle it subtends at any point on the remaining part of the
circumference.
2. The angle in a semi circle is a right angle
3. Angles in the same segment of a circle and subtended by the same arc
are equal.
4. The opposite angles of a cyclic quadrilateral are supplementary ( adds
up to 1800)
5. The exterior angle of a cyclic quadrilateral is equal to the interior
opposite angle.
6. The tangent of a circle is perpendicular (900) to the radius of that
circle at the point of contact.
7. The lengths of two tangents from an external point to the points of
contact on the circle are equal.
8. The angle between a tangent to a circle and a chord through the point
of contact is equal to the angle in the alternate segment.
9.
The line joining the centre of a circle to the midpoint of a chord is
perpendicular (900) to the chord.
Solution
a.
circle)
TPQ = 900 ( angles in a semi
b.
MTQ = 900( tangent is
perpendicular to the radius)
c.
TQS = 230 ( the angle between a
tangent and a chord is equal to the angle in the
alternate segment).
d.
SRQ=1130 ( opposite angles is a
cyclic quadrilateral are supplementary.
Since angle STO = 90 – 23 = 670
→
SRQ = 180 – 67 = 1130
Solution
a. VZW = 510 (the angle between a
tangent and a chord is equal to
the angle in the alternate
segment).
b. XYZ = 780
Since VWZ= 780 and ZWX= 180 –
78 = 1020 ( angles on a straight line)
→ XYZ= 180 – 102 =780
( opposite angles is a cyclic
quadrilateral are supplementary).
Solution
i.
ii.
iii.
iv.
ACD= 900 ( angle in a
semicircle)
EAD = 300 ( the angle between a
tangent and a chord is equal to
the angle in the alternate
segment.
EOD = 600 ( equal to two angles
at the circumference)
or
Since angle 0ED= 90 – 30 = 600
An angle ODE = 600 ( isosceles)
triangle
Angle EOD = 180 – 60 – 60 =
600
BCD = 1200 ( opposite angles in
a cyclic quadrilateral are
supplementary)
Since BED = 600
Then angle BCD = 180 – 60 =
1200
Solution
i.
PTS = 640
Since OSP = 260 (alternate angles)
Since OPS is an isosceles triangle
then OPS= 260
POS = 180 – 26 – 26 = 1280 (angles in
a triangle
128
→ PTS = 2 = 640 ( the angle at the
centre is twice the angle at the
circumference)
Solution
i.
Since angle BCD = 900
(angles in a semicircle)
Since triangle CBD = 450
Then angle
x = 180 – 90 – 45 = 450
ii.
The diagram ABCD is a
square all the angles are 900
Since the triangles are
isosceles and similar then all
the sides are equal.
Solution
i.
ii.
iii.
WQS = 500 ( the angle between a
tangent and a chord is equal to
the angle in the alternate
segment)
Triangle STQ is an isosceles
triangle. Since angle STQ = 1100
180−110
→ TSQ = TQS =
= 350
2
(the lengths of two tangent from
an external point to the points of
contact on the circle are equal.
→since SQT = 350 then angle
SWQ = 350
Then WSQ = 180 – 35 – 50 = 950
(Angles in a triangle)
iv.
WQY = 350 (alternate angles)
v.
WXY = 1450 ( opposite angles is
a cyclic quadrilateral are
supplementary)
Solution
BTP = 900( angle formed with a
tangent and a radius)
BAT = 900 ( angle in a
semicircle)
ABT = 500
Since ∆ BTP is a right angle
triangle then
ABT = 180 – 40 – 90 = 500
ACT= 500(angles in the same
segment are equal.)
Hence the area of the shades region is
35.4 – 21 = 14.4cm2
In the diagram below, NOT DRAWN TO
SCALE, PQ is a tangent to the circle PTSR,
so that angle RPQ= 460 and angle RQ
P = 320 and TRQ is a straight line.
Calculate, giving a reason for each
step of your answer
i. Angle PTR
Solution
a. MCD = 480
Since ∆OCD is an isosceles triangle then
180−96
angle OCD= 2 = 420
Since angle OCM= 900 ( angle form witha
tangent and a radius)
Then angle MCD = 90 – 42 = 480
ii. CMD = 840 ( the lengths of two tangents
from an external point to points on the circle
are equal)
hence the triangle MCD is isosceles
→ 180 - 48 – 48 = 840
b. Area = ½ cdsinO
= ½ x 6.5 x 6.5 x sin 96
= 21cm2
ii. the area of the sector OCD
𝜃
96
= 360 𝜋𝑟 2 = 360 × 3.14 × (6.5)2 = 35.4
ii.
iii.
Angle TPR
Angle TSR