Question

A circle with centre p has a chord BQ=24cm.the diameter of the circle is 26 cm.Find that distance of chord from the centre of circle

Answer 1

Answer:

5 cm

Step-by-step explanation:

Answer

R.E.F image

Given diameter of circle = 26 cm

diameter =2r = 26 cm

radius = 13 cm

length of chord of circle = 24 cm

Given OA = 13 cm

AC=

2

AB

=12cm

∴OC=

OA

2

−AC

2

(By Pythagoras theorem)

=

(13)

2

−(12)

2

=

169−144

=

25

=5cm

Answer 2

$\\ \bigstar \: \large \bold{ \underline{ \underline{Given :- }}}$

• A circle with centre p.
• Length of Diameter = 26 cm
• Length of chord = BQ = 24 cm
• Line PC is perpendicular to chord AB.

$\\ \bigstar \: \large \bold{ \underline{ \underline{To \: find :- }}}$

• Distance of chord from the centre of circle. ( Distance PC )

$\\ \bigstar \: \large \bold{ \underline{ \underline{Solution :- }}}$

At first, We have to find Th radius of circle

So,

$\\ \sf \: \: \: \: Diameter = {Radius} \times 2$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: {Radius} = \frac{Diameter}{2}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: {Radius} = \frac{26}{2}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \green{{Radius} = PA = 13 }$

So, $\sf AC = \dfrac{AB}{2} = 12cm$

Now, Using Pythagoras theorem .

$\\ \: \: \: \: \sf \: PC^{2} = PA^{2} - AC ^{2}$

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: = 13^{2} - 12^{2}$

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: = 169- 144$

$\\ \: \: \: \: \sf \: \: \: \: \: \: \: \: = 25$

Therefore, $\sf PC = \sqrt{25}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \red{\boxed{\frak{ PC = 5cm}} }$

Hence, The distance of chord from the centre of circle is 5 cm .