Question

AB is the chord of a circle with centre o of AB=12 and op =8 cm find the radius of the circle​

Answer 1

Given :

• AB = 12 cm
• OP = 8 cm

To find :

• Radius of the circle

According to the question,

By using Pythagoras theorem,

$\sf : \implies{OB {}^{2} = OP {}^{2} + PB {}^{2} }$

$\sf : \implies{OB {}^{2} = {(8 \: cm)}^{2} + {(6 \: cm)}^{2} }$

$\sf : \implies{OB { }^{2} = 64 \: {cm}^{2} + 36 \: {cm}^{2} }$

$\sf: \implies{OB {}^{2} = {100 \: cm}^{2} }$

$\sf: \implies{OB = \sqrt{100 \: cm {}^{2} } }$

${ \underline{ \boxed{\sf { \pink{ : \implies{OB = 10 \: cm}}}}}} \: \bigstar$

$\therefore{ \underline{ \sf{So, the \: radius \: of \: the \: circle \: is { \textsf{ \textbf{10 cm.}}}}}}$

_________________

Theorem 1 :

The angle subtended by an arc of a circle at the centre is doubled the angle subtended by it at any point on the remaining part of the circle.

Theorem 2 :

The angle in a semicircle is a right angle.

Theorem 3 :

Angles in the same segment of a circle are equal.

Answer 2

Answer:

Given :-

• AB is a chord of a circle
• Centre as O
• AB=12
• OP = 8cm

To Find :-

Radius

Solution :-

Apply Pythagoras theorem

$\tt \implies \: OB {}^{2} = OP {}^{2} + PB {}^{2}$

$\tt \implies \: OB {}^{2} = {8}^{2} + {6}^{2}$

$\tt \implies \: OB {}^{2} = 64 + 36$

$\tt \implies \: OB {}^{2} = 100$

$\tt \implies \: {OB}^{} = \sqrt{100}$

${\boxed {\orange{\underline{\frak {OB = 10}}}}}$

Radius is 10 cm.