Question

$\large \boxed{\frak{question}}$


A chord of length 48cm is drawn in a circle of radius 25cm. Calculate its distance from the centre of the circle.​

Answer 1

$\huge\underline\mathfrak\pink{\:♡Solution\:♡}$

Answer :-

x = 7 cm

Explanation :-

Given :-

l = 48 cm

r = 25 cm

Solution :-

Distance of chord from centre of the circle is -

r^2 = (l/2)^2 + x^2

25^2 = (48/2)^2 + x^2

25^2 = 24^2 + x^2

625 = 576 + x^2

x^2 = 625 - 576

x^2 = 49

x = 7 cm

Hence, distance of chord from the centre is 7 cm.

Hope this helps you...

Answer 2

Answer:

$\large{{\boxed{\pink{\sf{Answer = 7cm}}}}}$

Step-by-step explanation:

Given :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf where \begin{cases} & \sf{radius \: of \: the \: circle = 25cm} \\ \\ & \sf{length \: of \: the \: chord = 48cm} \end{cases}\end{gathered} \end{gathered} \end{gathered}\end{gathered}$

To Find :

• the distance from the centre of the circle to the chord

Solution :

$\underline \frak{ \dag \: as \: we \: know}$

$\boxed{ \pink{ \sf{Pythagoras \: Theorem =AC² = AB² + BC²}}}$

__________________________________________

According to the image above.

ㅤ

AB is the length of the cord

OM is perpendicular to AB

ㅤ

OA = 25cm

OM ⊥ AB

ㅤ

M is the mid-point of AB

$\sf \: AM = \frac{1}{2} AB$

$\sf \: AM = \frac{1}{2} \times 48$

$\sf \: AM = 24cm$

ㅤ

∆OAM is a right angle triangle

ㅤ

By Using Pythagoras theorem

$\\ \sf \: OA² = OM² + AM²$

$\\ \sf \: 25² = OM² + 24²$

$\\ \sf \: OM² = 25² - 24²$

$\\ \sf \: OM = \sqrt{625 - 576}$

$\\ \sf \:OM = \sqrt{49}$

$\\ \sf \: OM = 7cm$

__________________________________________

$\\ \underline \text{Therefore, the distance from the centre of the circle to the chord is 7cm}$