In a circle of radius 6cm,a chord is drawn 3cm from the centre of the circle. find the length of the minor arc cut off the chord and draw a diagram of the question.
Here is the diagram of the question:
```
O
/ \
/ \
/ \
/ 3 \
/ \
/ \
A --- 6cm --- B
```
We want to find the length of the minor arc AB.
First, we need to find the length of the chord AB. To do this, we can use the Pythagorean theorem. We have a right triangle OAB, where OA = OB = 6cm and AB = 2 x 3cm = 6cm (since the chord is 3cm from the center, and the full diameter is 2 x the radius). So we have:
```
AB^2 = OA^2 - OB^2
AB^2 = 6^2 - 6^2
AB^2 = 36 - 36
AB^2 = 0
```
So AB = 0. This is not possible, since we know that AB is a non-zero chord. What went wrong?
We made an error when we assumed that OAB was a right triangle. In fact, it is an isosceles triangle, since OA = OB. This means that the altitude from O bisects AB, and we have two right triangles OAP and OBP, where AP = BP = 3cm and OA = OB = 6cm. Again using the Pythagorean theorem, we have:
```
AP^2 = OA^2 - OP^2
AP^2 = 6^2 - 3^2
AP^2 = 27
AP = sqrt(27) = 3 sqrt(3) cm
So AB = 2AP = 6 sqrt(3) cm.
```
Now we can find the length of the minor arc AB. Since the central angle subtended by AB is 120 degrees (it is one third of the full circle), and the radius is 6cm, we have:
```
arc length = (central angle/360) x 2πr
arc length = (120/360) x 2π(6)
arc length = (1/3) x π(6)
arc length = 2π cm
```
So the length of the minor arc AB is 2π cm (or approximately 6.28 cm).