A number of circles touch each ather. The area of the smallest circle is 4picm2 and each consecutive circle has area 9/4 times that of the previous one. If the distance AB=665/8. Line AB passes at the centre of the circles. How many circles are there?

The smallest circle has a radius of 2.

Each circle's radius is 3/2 that of the previous one.

If there are n circles, then since the first and last circles are only traversed halfway, we have

2 + 4(3/2) + 4(3/2)^2 + ... + 2(3/2)^(n-1) = 665/8

Hmmm. I get a non-integer solution for n.

Below are the lengths of AB as the number of circles increases
2
4*3/2 = 6; 2+6 = 8
6*3/2 = 9; 8+9 = 17
9 * 3/2 = 27/2; 17 + 27/2 = 61/2
and so on.

Have I misread something?

it saids 423+5123

To determine the number of circles in this scenario, we can use the formula for the area of a circle, which is A = πr^2, where A is the area and r is the radius of the circle.

Given that the area of the smallest circle is 4π cm^2, we can find the radius of this circle by rearranging the formula:

4π = πr^2
Dividing both sides by π: 4 = r^2
Taking the square root of both sides: r = 2 cm

Now, we know that the area of each consecutive circle is (9/4) times that of the previous one. Starting from the smallest circle, we can calculate the areas of the subsequent circles by multiplying the previous area by (9/4).

Let's denote the areas of the circles as A1, A2, A3, and so on, with A1 being the area of the smallest circle.

A1 = 4π cm^2 (given)
A2 = (9/4) * A1
A3 = (9/4) * A2
A4 = (9/4) * A3

To find the radius of each circle, we can substitute the area values into the area formula:

A = πr^2

For A1: 4π = πr1^2
Simplifying: 4 = r1^2
Taking the square root: r1 = 2 cm (same as before)

For A2: (9/4) * A1 = πr2^2
Substituting the value of A1: (9/4) * 4π = πr2^2
Simplifying: 9 = r2^2
Taking the square root: r2 = 3 cm

For A3: (9/4) * A2 = πr3^2
Substituting the value of A2: (9/4) * [(9/4) * 4π] = πr3^2
Simplifying: (?)