the radius of a circle is 8cm and the length of one of its chords is 12cm. find the distance of the chord from the centre

If you join the centre of the circle with each end of the chord, you have an isosceles triangle with equal legs of 8cm and base 12cm.

The altitude h(distance of centre from chord) cuts the triangle into two right triangles with half base = 6, and hypotenuse 8.

Using Pythagoras theorem,
h²=8^2-6²=28
h=√28

I CANT PLS HELP ME

using Pythagoras theorem

half of chord from center = 12/2 = 6 cm.
8square= x square + 6 square
64 = x+36
x square = 64-36
x square = 28
x = 5.291 cm

To find the distance of the chord from the center of the circle, we can use the following formula:

Distance of chord from center = √(r^2 - (l/2)^2)

Where:
- r is the radius of the circle
- l is the length of the chord

In this case, the radius (r) is 8cm and the length of the chord (l) is 12cm. Let's plug these values into the formula:

Distance of chord from center = √(8^2 - (12/2)^2)
= √(64 - 36)
= √28
≈ 5.29 cm

Therefore, the distance of the chord from the center is approximately 5.29 cm.