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​

Answer 1

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.

$\rule{150}2$

Answer 2

$\huge\bold\green{Question}$

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