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

In the figure\(\angle \)POQ = 100⁰
\(\angle \)PQR = 30⁰
\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\times\)100 = 50⁰
In\(\triangle \)OPQ = \(\angle \)OQP =\(\angle \)POQ = 180⁰
2\(\angle \)OPQ = 180⁰
2\(\angle \) OPQ = 80⁰
\(\therefore \) \(\angle \)OPQ = 40⁰ 
In \(\triangle \)PRQ,
\(\angle \)R+\(\angle \)P+\(\angle \)Q = 180⁰
50⁰ + (40 + x) + 30⁰ = 180⁰
(40 + x)⁰ = 180⁰ - 80 = 100⁰
x⁰ = 100⁰ - 40⁰ = 60⁰
\(\therefore \) \(\angle \)RPO = x = 60⁰

If CD 2 cm and R is the radius, i.e. OC = OA = OB = R
\(\therefore\) OD = OC - DC = R - 2 cm
since AB = 6 cm
\(\Rightarrow \) AD = DB = \(\cfrac { 6 }{ 2 } \) = 3 cm
In \(\triangle \)OBD,

\(\Rightarrow \) OB² = OD² + BD² (By Pythagoras theorem)
\(\Rightarrow \) R² = (R - 2)² + 3²

\(\Rightarrow \) 4R = 13
\(\Rightarrow \)  R = \(\cfrac { 13 }{ 4 } \) = 3.25 cm

\(\angle\)DBC = 40⁰
\(\angle\)BAC = 60⁰
(i) \(\angle\)CAD = 60⁰
(\(\therefore \) angles in the same segment are equal)
(ii) \(\angle\)BCD+\(\angle\)BAD = 180⁰
(\(\therefore\) In cyclic quadrilateral opposite angles are supplementary)

\(\angle\)BCD+ (\(\angle\)BAD +\(\angle\)CAD) =180⁰
\(\angle\)BCD+ (60⁰ + 40⁰) = 180⁰
\(\angle\)BCD = 180⁰-100⁰ = 80⁰

OD = DP =\(\cfrac { 4cm }{ 2 } \) = 2 cm
AD = BD =\(\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } \)
= \(\sqrt { 25-19 } \)
= \(\sqrt { 9 } \)
= 3cm
\(\therefore\) The length of the common chrod AB = AD + BD = (3 + 3) cm = 6 cm

BD = \(\sqrt { { 15 }^{ 2 }+{ 12 }^{ 2 } } \)
= \(\sqrt { 225-144 } \)
= \(\sqrt { 81 } \)
= 9 cm
\(\therefore\) length of the chord AB = 9 + 9 = 18

OA² = AC² + OC²
Radius of the circle = \(\sqrt { { 8 }^{ 2 }+{ 15 }^{ 2 } } =\sqrt { 64+225 } \)
= \(\sqrt { 289 } =17cm\)

\(=\sqrt { \left( 25cm \right) ^{ 2 }-\left( 20\quad cm \right) ^{ 2 } } \)
\(=\sqrt { \left( 625-400 \right) cm } \)
\(=\sqrt { 225 } =15cm\)

Given ㄥABC = 30⁰
ㄥAOC = 2ㄥABC
(The angle subtended by an arc at the centre is double the angle at any point on the circle)
= 2 x 30⁰
= 60⁰

In PQR,
PQ = 5 cm,
PR = 6 cm
\(\angle \)QPR = 60°

Construction :
Step 1: Draw △ PQR with the given measurement
Step 2: Draw perpendicular bisectors of any two sides (PQ and QR) to find the mid points of PQ and QR.
Step 3: Draw medians PD and RE. Let them meet at G.
Step 4: G is the centroid of the given △PQR.

In \(\triangle \)LMN
LM = 7.5 cm,
MN = 5 cm,
LN = 8 cm

Construction :
Step 1: Draw \(\triangle \)LMN with LNM = 8 cm, MN = 5 cm, LM = 7.5 cm
Step 2: Construct perpendicular bisectors for any two sides (LN and MN) to find the mid points of LM and MN.
Step 3: Draw the medians LD, ME. Let them meet at G.
Step 4: G is the centroid of the triangle LMN.

Construction :
Step 1: Draw \(\triangle \)ABC with BC = 8 cm. AC = 10 cm with right angle at B.
Step 2: Construct angle bisectors of any two angles (B and C) and let them meet at I. I is the incentre.
Step 3: Draw perpendicular from I to any side of the triangle to meet BC at D.
Step 4: With I as centre, ID as radius draw the incircle, which touches all the three sides of the triangle internally. In radius = 1.9 cm.

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