Question

what is the length of the perpendicular from the centre of a circle of radius 5 cm on chord of circle of length 8 cm

Answer 1

So let's say there's a circle with center O as shown in the diagram above⬆️.
With a chord AB of length 8cm.

Now,
You might know that a perpendicular from the center of the circle falling on the chord bisects it.
Therefore,
AC= CB

As you can see that,
AC + CB = AB
AC + AC = AB
2AC = AB = 8cm

AC = 4cm

As the circle has the radius of 5
Therefore,

OA = 5 cm

Now,
By using Pythagoras theroem in ΔOAC

${AC}^{2} + {OC}^{2} = {OA}^{2}$
Substitute the values you have and you'll end up with,

16 + OC^2 = 25
OC^2 = 25-16
OC = √9=3

Therefore the required length of the perpendicular here is 3cm.

Answer 2

Answer: chord AB of length 8cm.

Now,

You might know that a perpendicular from the center of the circle falling on the chord bisects it.

Therefore,

AC= CB

As you can see that,

AC + CB = AB

AC + AC = AB

2AC = AB = 8cm

AC = 4cm

As the circle has the radius of 5

Therefore,

OA = 5 cm

Now,

By using Pythagoras theroem in ΔOAC

Substitute the values you have and you'll end up with,

16 + OC^2 = 25

OC^2 = 25-16

OC = √9=3

Therefore the required length of the perpendicular here is 3cm.