110°

50°, 80°

In \(\Delta ABC\),
\(\angle BAC+\angle ABC+\angle ACB=180°\)
Sum of all the angles of a triangle is 180°
⇒ \(\angle BAC+69°+31°=180°\)
⇒ \(\angle BAC+100°=180°\)
⇒ ​​​​​​ \(\angle BAC=180°-100°=80°\)  .........(1)
Now, \(\angle BDC=\angle BAC\)
Angles in the same segment of a circle are equal = 80°.       Using (1)

Let BD = x m

Then in right triangle ODB,
OB² = OD² + BD²
By Pythagoras Theorem
⇒ (20)² = OD² + x²
⇒ OD²=  400 - x²⇒ OD =\(\sqrt { 400-x^{ 2 } } \)
Again, area of equilateral triangle ABC
= Area of \(\Delta OBC\) + Area of \(\Delta OCA\) + Area of \(\Delta OAB\)
= 3 Area of \(\Delta OBC\)=  3\(\frac { \left( BC \right) \left( OD \right) }{ 2 } \)
\(=3x\sqrt { 400-x^{ 2 } } \)                       ....(2)
⇒  \(\sqrt { 3 } \sqrt { 400-x^{ 2 } } =x\)
Squaring both sides,
3(400 - x²) = x²
⇒ 1200 - 3x² = x²
⇒  4x² = 1200   ⇒ x² = 300
⇒ \(x=10\sqrt { 3 } \)  ⇒ BD=\(10\sqrt { 3 } \)
⇒  \(2BD=20\sqrt { 3 } \)  ⇒ \(BC=20\sqrt { 3 } \)
Hence, the length of string of each phone is \(20\sqrt { 3 } \) m.

(i) The centre of circle lies in interior of the circle.
(ii) A point distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and the chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.

Since chord AB and chord CD subtend equal angles at the centre,
i.e. ∠AOB = ∠COD [Each = 60°]
∴ Chord AB = Chord CD
⇒ Chord AB = 5 cm [∴ Chord CD = 5 cm]
Thus, the required lenth of chord AB is 5 cm.

Given AB and CD are the chords of a circle with centre at O such that AB = CD

To Prove: \(\angle\)AOB = \(\angle\)COD
Proof: In \(\Delta\)AOB and \(\Delta\)COD
AO = CO        (radii of same circle)
AB = CD         (given)
BO = DO        (radii of same circle)
\(\Delta\)AOB\(\cong \) \(\Delta\)COD   (SSS)
\(\angle\)AOB = \(\angle\)COD  (c.p.c.t.) 2 Hence Proved.

(i) True. Because all points on the circle are equidistant from the centre of the circle and this equal distance is called radius of the circle.
(ii) False. Because in circle, infinitely many equal chordscan be drawn.
(iii) False. Because all three arcs are equal, so there is no difference berween the major and minor arcs.
(iv) True. By the definition of diameter, it is rwice the radius of a circle.
(v) False. Because the sector is the region between two radii and an arc.
(vi) True. Because circle is a rwo dimensional figure and it can also be referred as plane figure.

40°, 40°

The perpendicular from the center bisects the chord.

\(\Rightarrow\) DN =1/2 CD = 6cm
and BM = x,
OD =OB      (Radius of the same circle)
OD² = OB²
\(\Rightarrow\) ON² + ND² = OM² + MB²
\(\Rightarrow\) 8² + 6² = 6² + x²
\(\Rightarrow\) x = 8 cm
\(\Rightarrow\) BM =8
\(\Rightarrow\) AB = 2BM
= 2x8
\(\Rightarrow\) AB = 16 cm.

6 cm

(i) x = 2 x 35° = 70°
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(ii) \(x=\frac { 1 }{ 2 } \times 110°=55°\)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iii) \(x=\frac { 1 }{ 2 } \times 70°=35°\)
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iv) x = 180° - (90° + 55°) I ∵ Angle in a semi-circle is 90°
= 180° - 145° = 35°
(v) \(x=\frac { 1 }{ 2 } \times \left( 180°-120° \right) \)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle \(=\frac { 1 }{ 2 } \left( 60° \right) =30°\). 

Given: In figure, AB and CD are equal chords of a circle whose centre is O. OM 丄 AB  and ON 丄 CD.
To Prove: \(\angle OMN=\angle ONM\).
Proof: ∵ Chord AB = Chord CD
 ∴ OM=ON     .........(1)
 | Equal chords of a circle are equidistant from the centre of the circle
In \(\Delta OMN\),
OM=ON     I From (1)
∴ \(\angle OMN=\angle ONM\) | Angles opposite to equal sides of a triangle are equal.

\(AC=\sqrt { OA^{ 2 }+OC^{ 2 } }\) 
\(\quad =\sqrt { 10^{ 2 }-8^{ 2 } } =6\quad cm\) 
\(\therefore AB=2AC=12\quad cm\)

\(AC=BC=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } (24)=12\quad cm\)

\(OA=\sqrt { OC^{ 2 }+AC^{ 2 } } \)
\(=\sqrt { 5^{ 2 }+12^{ 2 } } =13\quad cm\) 

Theorem

\(\Delta AOB\cong \Delta COD\)

See a circle

AB is a line segment, which subtends equal angles at two points C and D. That is
\(\angle\) ACB = \(\angle\) ADB
To show that the points A, B, C and D lie on a circle let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD (or extended AD) at a point, say E (or E').
If points A, C, E and B lie on a circle,
\(\angle\) ACB = \(\angle\) AEB
But it is given that \(\angle\) ACB = \(\angle\) ADB.
Therefore, \(\angle\) AEB = \(\angle\) ADB.
This is not possible unless E coincides with D.
Similarly, E' should also coincide with D.

(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.