Math, asked by akshita46167, 9 months ago

Radius of a circle with centre O is 25 cm find the distance of a chord from the centre if length of the chord is 48cm​

Answers

Answered by ShírIey
171

AnswEr:-

Required Distance is 7cm.

Step by Step Explanation:-

Given:-

  • A Circle with Center 'O'.
  • Radius of Circle = 25 cm.
  • Length of the Circle= 48cm.

To find :-

Distance of Chord

Finding:-

:\implies\sf\; AB =  \dfrac{BC}{2}

:\implies\sf\; AB = \dfrac{48}{2}

:\implies\sf\; AB = 24 cm

:\implies\sf\; OB = 25 cm

\rule{150}2

Now, In ∆ OAB

By Using Pythagoras theorem:-

:\implies\sf\; OB^2 = OA^2 + AB^2

:\implies\sf\; 25^2 = OA^2 + 24^2

:\implies\sf\; OA^2 = (25 + 4) (24 - 4)

:\implies\sf\; OA^2 = 49

:\implies\sf\; OA = \sqrt{49}

:\implies\large\boxed{\sf{\red{OA = 7cm}}}

This is the required Distance.

\rule{150}2

Attachments:
Answered by Anonymous
6

\huge\bold\green{Question}

Radius of a circle with centre O is 25 cm find the distance of a chord from the centre if length of the chord is 48cm

\huge\bold\green{Answer}

●According to question Given data is :-

→ Circle with Center 'O'

→ Radius of Circle = 25 cm.

→ Length of the chord of circle = 48cm.

●We have to find :-

Distance of the Chord

\tt {AB = \dfrac{BC}{2}}

\tt { AB = \dfrac{48}{2}}

\tt { AB = 24 cm}

\tt {OB = 25 cm}

So, In ∆ OAB

Simply by Using Pythagoras Theorem:-

\tt{OB^2 = OA^2 + AB^2}

\tt{25^2 = OA^2 + 24^2}

\tt{OA^2 = (25 + 4) (24 - 4)}

\tt{OA^2 = 49}

\tt{OA = \sqrt{49}}

= \tt\green{OA = 7cm}

Hence, the required distance of chord is 7 cm

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