In the figure OA = OD = 6cm   (\(\therefore\) radius = 6 cm)
AB = 8cm (chord)
CD = 10 cm (chord)
OM = \(\sqrt { { 6 }^{ 2 }-{ 4 }^{ 2 } } \)   ( \(\because \) OM bisects the chord and \({ \bot }^{ r }\) to the chord)
= \(\sqrt { 36-16 } =\sqrt { 20 } cm\)
ON=\(\sqrt { { 6 }^{ 2 }-{ 5 }^{ 2 } } =\sqrt { 36-25 } =\sqrt { 11 } cm\)
ONPM is a rectangle with all the angles 90° and with length \(\sqrt { 20 } \) cm, breadth \(\sqrt { 11 } \) cm.
We need to find OP which is the diagonal of the rectangle ONPM

\(\therefore \ OP=\sqrt { { ON }^{ 2 }+{ NP }^{ 2 } } =\sqrt { \sqrt { { 11 }^{ 2 }+\sqrt { 20^{ 2 } } } } \)
(\(\therefore \) OM = NP, opposite sides of the rectangle)
\(=\sqrt { 11+20 } =\sqrt { 31 } =5.56cm=5.6cm\)

In this figure

OE = \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } =\sqrt { 169-25 } =\sqrt { 144 } =12cm\)
OF = \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } =\sqrt { 169-25 } =\sqrt { 25 } =5cm\)
The shortest distance between = 12 + 5 = 17 cm

In the cyclic quadrilateral \(\angle\)A + \(\angle\)C = 180⁰

6y = 180⁰
\(y=\cfrac { { 180 }^{ 0 } }{ 6 } ={ 30 }^{ 0 }\)
\(\angle\)B+\(\angle\)D = 6\(\times\)-4+7\(\times\)+2
13\(\times\)-2 = 180⁰
13\(\times\) = 180 + 2 = 182⁰
x = \(\cfrac{182}{13}={14}^{0}\)
\(\therefore\) \(\angle\)A = 2(30) + 4⁰ = 64⁰
\(\angle\)B = 6(14) - 4⁰ = 84 - 4 = 80⁰
\(\angle\)C = 4(30) - 4 = 120 - 4 = 116⁰
\(\angle\)D = 7(14) + 2 = 98 + 2 = 100⁰

(i) \(\angle\)ACD = 180⁰-(90⁰+ 35⁰)
= 18⁰-125⁰ = 55⁰
(ii)\(\angle\)ACB = 180⁰ - (90⁰ + 40⁰)
= 180⁰-130⁰ = 50⁰
(iii)\(\angle\)ADC = 90⁰
\(\angle\)CAE = 180⁰-120⁰ = 60⁰
\(\therefore\) \(\angle\)DAE = 60⁰- 35⁰ = 25⁰

The distance between two chords FE
= OE + OF
OE = \(\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } } =\sqrt { 100-36 } =\sqrt { 64 } =8cm\)
OF = \(\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } } \)
= \(\sqrt { 100-64 } \)
= \(\sqrt { 36 } =6cm\)
\(\therefore \) FE = 8 cm + 6 cm = 14 cm
\(\therefore\) Distance between the chords is 14 cm

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

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

In ΔABC, AB = BC = 6 cm, ㄥB = 80°.

Construction:
Step 1: Draw \(\triangle \)ABC with BC = 6 cm. AB = 6 cm, AB = 6 cm, and ㄥB = 80°.
Step 2: Construct the incentre I and ID is the inradius, as in the previous sums.
Step 3: Draw incircle with I as centre and ID as radius. It touches all the three sides internally.
Step 4: Measure inradius. Inradius = 1.7 cm

Construction :
Step 1: Draw\(\triangle \)ABC with AB = 9 cm. ㄥA = 115°,ㄥB = 40°.
Step 2: Construct angle bisectors of any two angles (B and C). Let them meet at I. I is the incentre of \(\triangle \)ABC.
Step 3: Draw perpendicular from I to any side (BC) to meet BC at D.
Step 4: Draw incircle, with I as centre and ID a radius. Measure the inradius.

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. Inradius = 1.9 cm.