(i) Let us assume that A, Band C are the position of Priyanka, Sania and David respectively on the boundary of circular park with centre O.
Draw AD\(\bot\) BC
Since the centre of the circle coincides with the centroid of the equilateral \(\triangle\) ABC
\(\therefore\) Radius of circumscribed circle = \(\frac{2}{3}\) AD
\(\Rightarrow 20=\frac { 2 }{ 3 } AD\)
\(\Rightarrow AD=20\times \frac { 2 }{ 3 } \)
\(\Rightarrow AD=30m\)
Now, AD\(\bot\)BC, and let AB=BC=CA=x
\(\Rightarrow BD=CD=\frac { 1 }{ 2 } BC=\frac { x }{ 2 } \)

In rt. \(\triangle\)BDA, D=90⁰
By Pythagoras Theorem, we have
AB²=BD²+AD²
\(\Rightarrow { x }^{ 2 }={ \left( \frac { x }{ 2 } \right) }^{ 2 }+{ (30) }^{ 2 }\)
\(\Rightarrow{ x }^{ 2 }-{ \frac { { x }^{ 2 } }{ 4 } }=90\)
\(\Rightarrow \frac { 3 }{ 4 } { x }^{ 2 }=90\)
\(\Rightarrow { x }^{ 2 }=900\times \frac { 4 }{ 3 } \)
\(\Rightarrow { x }^{ 2 }=1200\)
\(\therefore x=\sqrt { 1200 } =20\sqrt { 3 } \)
Hence, the distance between each of them is \(20\sqrt { 3 } .\)
(ii) Properties of the circle, equilateral triangle and Pythagoras theorem.
(iii) Live and let live.

(i) Given: A circle C(O, r) and chord AB = chord AC. AD is bisector of \(\angle\)CAB.
To Prove: Centre O lies on the bisector of \(\angle\)BAC
Construction Join Be, meeting bisector AD of \(\angle\)BAC, at M.

Proof: In triangles BAM and CAM,
AB=AC    (given)
\(\angle\)BAM= \(\angle\)CAM  (given)
AM=AM   (Common)
\(​​\triangle BAM\cong \triangle CAM\)   (SAS)
\(\Rightarrow\)  BM=CM
and \(\angle\)BMA = \(\angle\)CMA
As \(\angle\)BMA + \(\angle\)CMA = 180⁰ (linear pair)
\(\Rightarrow\) \(\angle\)BMA = \(\angle\)CMA = 90⁰
\(\Rightarrow\) AM is the perpendicular bisector of the chord BC
\(\Rightarrow\) AM passes through the centre O.
[\(\because\) Perpendicular bisector of chord of a circle passes through the centre of the circle]
Hence, the centre of the park lies on the angle bisector of \(\angle\)BAC
(ii) Congruency of triangles by SAS axiom (Geometry)
(iii) Cleanliness and respect for labour.

(i) AB and CD intersect atO0
\(\therefore\) \(\angle\)AOD = \(\angle\)BOC
(Vertically opp. angles) ...(i)
In \(\triangle\)AOD and \(\triangle\)BOC, we have 
\(\angle\)AOD = \(\angle\)BOC ...(ii)
\(\angle\)DAO = \(\angle\)CBO = 90° (Given)
and AD = BC (Given)
\(\triangle AOD\cong \triangle BOC\)
(By AAS congruence criterion)
\(\Rightarrow\) OA = OB (By c.p.c.t.)
i.e., O is the mid-point of AB
Hence, CD bisects AB.
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

In \(\triangle\)ABC, as AB is the greatest side
\(\Rightarrow\) AB > BC \(\Rightarrow\) \(\angle\)C > \(\angle\)A
AB > AC \(\Rightarrow\) \(\angle\)C > \(\angle\)B
On adding (1) and (2), we get
2\(\angle\)C > \(\angle\)A + \(\angle\)B
\(\Rightarrow\) 2\(\angle\)C + \(\angle\)C > \(\angle\)A + \(\angle\)B + \(\angle\)C
\(\Rightarrow\) 3\(\angle\)C > 180°
\(\therefore\) \(\angle\)C > 60°.

(i) Required number of students = 8 + 32 = 40
(ii) Here, We notice that classes are continuous but class-size is not the same for all the classes. We notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of rectangle (adjusted frequency) for each class.
Length of rectangle (adjusted frequency) =\(\frac { Frequency\ of\ Class }{ Width\ of\ class } \times Minimum\ class-size\)

Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies as heights.
Histogram representing marks obtained by students in unit test of Mathematics.

(iii) Hardwork and Dilligence.

Required histogram is as follows:

Proof: In \(\triangle\)ADE, we have
AD = AE
\(\angle\)ADE = \(\angle\)AED
180°- \(\angle\)ADE = 180°- \(\angle\)AED
\(\Rightarrow\) \(\angle\)ADB = \(\angle\)AEC
Consider \(\triangle\)ABD and \(\triangle\)ACE
AD =AE
\(\angle\)ADB = \(\angle\)AEC
BD = EC
By SAS congruence,
\(\triangle ADB\cong \triangle AEC\)
By c.p.c.t., AB = AC
\(\therefore\) \(\triangle\)ABC is an isosceles triangle.

ㄥR=80° (Given)
SR II PQ and RQ is a transversal
ஃ ㄥR+ㄥQ=180°
ㄥQ=180°-80°
= 100°
Similarly, ㄥQ+ㄥP=180°
⇒ ㄥP=180°-100°=80°
and ㄥS+ㄥR=180°
⇒ ㄥS=180°-80°=100°
Hence, ㄥP=80°, ㄥQ=100°, ㄥR=80°, ㄥS=100°

(ii) Property of co-interior angles when a pair of straight lines intersected by another straight line (Geometry)
(iii) Diligence i.e., dedication, determination, and hard work.

\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)is an irrational number.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } =\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 2-1 } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 1 } } =\sqrt { 2 } -1\)
which is an irrational number.
Let, there is a number x such that x³ is an irrational number but x⁵ is a rational number.
Let, x =\(\sqrt[5]{7}\) be the number.
⇒ x³ = (5√7)³ = (7)^3/5
is an irrational number.
But x⁵ = (\(\sqrt[5]{7}\))⁵=(7)^5/5  = 7
=7 is a rational number.
(ii) Accepting own mistakes gracefully, co-operative learning among the classmates.

\(={ \left( { x }^{ { a }^{ 2 }-{ b }^{ 2 } } \right) }^{ \frac { 1 }{ a+b } }.{ \left( { x }^{ { b }^{ 2 }-{ c }^{ 2 } } \right) }^{ \frac { 1 }{ b+c } }.{ \left( { x }^{ { c }^{ 2 }-{ a }^{ 2 } } \right) }^{ \frac { 1 }{ c+a } }\)
\(={ x }^{ \frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b } }.{ x }^{ \frac { { b }^{ 2 }-{ c }^{ 2 } }{ a+b } }.{ x }^{ \frac { { c }^{ 2 }-{ a }^{ 2 } }{ c+a } }\)
\(={ x }^{ a-b }.{ x }^{ b-c }.{ x }^{ c-a }\)
\(={ x }^{ 0 }\)