By the exterior angle property of a cyclic quadrilateral, we get, y°=100° and
x° + 30°= 60⁰ and so x° = 30°

PQRS is a cyclic quadrilateral
Given ㄥPSR = 70°
ㄥPSR+ㄥPQR = 180° (state reason________)
70° +ㄥPQR = 180°
ㄥPQR = 180°− 70°
ㄥPQR = 110°
In ΔPQR we have, ㄥPQR+ㄥPRQ+ㄥQPR = 180° (state reason_________)
110° +ㄥPRQ + 40° = 180°
ㄥPRQ = 180°− 150°
ㄥPRQ = 30°

(i)
  
\({ x }^{ 0 }=\cfrac { 1 }{ 2 } \angle BOC=\cfrac { 1 }{ 2 } \times \left( { 30 }^{ 0 }+{ 60 }^{ 0 } \right) =\cfrac { 1 }{ 2 } \times { 90 }^{ 0 }={ 45 }^{ 0 }\)
(ii)

\(\angle QPR=\cfrac { 1 }{ 2 } \angle QOR=\cfrac { 1 }{ 2 } \times { 80 }^{ 0 }={ 40 }^{ 0 }\)
In \(\triangle QPR\)
\(\angle R+\angle P+\angle Q={ 180 }^{ 0 }\)
\(\angle Q\) = 180^o-120^o= 60^o
\(\angle Q\) = x^o+ 50^o = 60^o
x^o = 60^o-50^o = 10^o
(iii)

\(\angle \) MPN = 90° (Angle subtended by the diameter is 90°)
\(\angle OMP+\angle OPM={ 180 }^{ 0 }-{ 110 }^{ 0 }={ 70 }^{ 0 }\)
\(\angle MPO=\cfrac { 70 }{ 2 } =35^{ 0 }\)
But \(\angle MPN={ 90 }^{ 0 }\)
\(\therefore\) \(\angle \)OPN = 90^o - 35^o = 55^o
(iv)

Central angle is twice that of angle sub tended on the circumference
\(\angle \)YOZ = 120^o\(\times\)2 = 40^o
\(\therefore\) x+\(\angle\)YOZ = 360^o
x^o = 360^o- 240^o = 120^o
(v)

\(\angle\)BOC = 360^o- 240^o = 120^o
\(\therefore\) \(\angle\)BAC = x^o=\(\cfrac { { 120 }^{ 0 } }{ 2 } \) = 60^o

Using the theorem the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of a circle.
(i) ㄥPOR = 2ㄥPQR

x° = 2\(\times\)50°
x° = 100°
(ii) \(\angle M N L=\frac{1}{2}\) Reflex ㄥMOL
\(=\frac{1}{2} \times 260^{\circ}\)
x° = 130°

(iii) XY is the diameter of the circle.
Therefore ㄥXZY = 90°
(Angle on a semi – circle)
In ΔXYZ
x° + 63° + 90° = 180°
x° = 27°


(iv) OA = OB = OC (Radii)
In ΔOAC,
ㄥOAC = ㄥOCA = 20⁰
In ΔOBC
ㄥOBC =ㄥOCB = 35⁰
(angles opposite to equal sides are equal)
ㄥACB =ㄥOCA + ㄥOCB
x⁰ = 20⁰ + 35⁰
x⁰ = 55⁰

Given: Chord AB of the outer circle cuts the inner circle at C and D.

To prove: AB − CD = 2AC
Construction: Draw OM ⊥ AB
Proof: Since, OM ⊥ AB (By construction)
Also, OM ⊥ CD
Therefore, AM = MB ... (1) (Perpendicular drawn from centre to chord bisect it)
CM = MD ... (2)
Now, AB – CD = 2AM–2CM
= 2(AM–CM) from (1) and (2)
AB – CD = 2AC

Let AB be the chord and C be the mid point of AB
Therefore, OC ⊥ AB
Join OA and OC.

OA is the radius
Given OC = 2\(\sqrt{11}\)cm and OA = 12cm
In a right ΔOAC,
using Pythagoras Theorem, we get,
AC² = OA² - OC²
= 12²-(2\(\sqrt{11}\))²
= 144 - 44
= 100cm
AC² = 100cm
AC = 10cm
Therefore, length of the chord AB = 2AC
= 2 \(\times\) 10 cm = 20 cm

Step 1: Draw ΔPQR using the given measurements PQ = 8cm QR = 6cm and RP = 7cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.

Step 2 : Draw the medians PN and RM and let them meet at G.
The point G is the centroid of the given ΔPQR.

Step 1 : Draw the ΔABC with AB = 6cm, ㄥB = 65° and AC = 7cm

Step 2 : Construct the angle bisectors of any two angles (A and B) and let them meet at I.
Then I is the incentre of ΔABC. Draw perpendicular from I to any one of the side (AB) to meet AB at D

Step 3: With I as centre and ID as radius draw the circle. This circle touches all the sides of the triangle internally

Step 4: Measure inradius 
In radius = 1.9 cm