**In the Figure KE is parallel to LF which is parallel to MG which is parallel to NH etc. Also DK: KL= 2:1

DL:LM = 2:1
DM: MN=2:1 etc
If KE = 2 cm , calculate the length of the 10th parallel line,rounded off to one decimal place.(N.B NH is the 4th parallel line)

Given that KE is parallel to LF, LF is parallel to MG, MG is parallel to NH, and so on.

We are also given that DK : KL = 2 : 1, DL : LM = 2 : 1, DM : MN = 2 : 1, and so on.

We need to find the length of the 10th parallel line, knowing that NH is the 4th parallel line.

To calculate the length of the 10th parallel line, we need to find the ratio of DK : KL for each line.

Since DK : KL = 2 : 1 for each consecutive line, we can use this ratio to calculate the length of the 10th parallel line.

Given that DK = 2 cm for the first line, we can calculate the length of the 10th parallel line as follows:

1st parallel line: DK = 2 cm
2nd parallel line: 2 * DK = 4 cm
3rd parallel line: 2 * (2 * DK) = 8 cm
4th parallel line: 2 * (2 * (2 * DK)) = 16 cm
5th parallel line: 2 * (2 * (2 * (2 * DK))) = 32 cm
6th parallel line: 2 * (2 * (2 * (2 * (2 * DK)))) = 64 cm
7th parallel line: 2 * (2 * (2 * (2 * (2 * (2 * DK))))) = 128 cm
8th parallel line: 2 * (2 * (2 * (2 * (2 * (2 * (2 * DK)))))) = 256 cm
9th parallel line: 2 * (2 * (2 * (2 * (2 * (2 * (2 * (2 * DK))))))) = 512 cm
10th parallel line: 2 * (2 * (2 * (2 * (2 * (2 * (2 * (2 * (2 * DK)))))))) = 1024 cm

Therefore, the length of the 10th parallel line is 1024 cm, rounded to one decimal place.

To solve this problem, we will use the given ratios to find the length of each line segment and then use that information to calculate the length of the 10th parallel line.

Given that the ratio DK:KL = 2:1, we can assume that DK has a length of 2x and KL has a length of x.

Since KE is parallel to LF, we can deduce that KE = LF. And since KE is given to be 2 cm, LF is also 2 cm.

Next, let's find the lengths of the line segments DL and LM using the ratio DL:LM = 2:1. Let DL have a length of 2y and LM have a length of y.

Since DL is parallel to KE, LF, and MG, we know that DL = KE = LF = MG. Therefore, DL has a length of 2 cm.

By the given ratio, we can deduce that DM has a length of 2z and MN has a length of z.

Using the information given, we can now calculate the length of NH, the 4th parallel line. Since NH is parallel to KE, LF, MG, and DK, we can use the ratios DK:KL and DL:LM to find NH.

NH = DK + KL + LF + MG + DL + LM
= 2x + x + 2 + 2 + 2y + y
= 3x + 3y + 4

Since NH is the 4th parallel line, we need to calculate the length of the 10th parallel line, which would be NK, keeping in mind the given ratios.

NK = NH + DM + MN
= (3x + 3y + 4) + (2z) + (z)
= 3x + 3y + 3z + 4

Now substitute the given values:
KE = 2 cm, DK:KL = 2:1

Since DK:KL = 2:1, we can let DK = 2k and KL = k.

Therefore, we have:
2k = 2x
k = x

Now we can substitute the value of k into the equation for NH:
NH = 3(2x) + 3(2y) + 4
= 6x + 6y + 4

Since NH is the 4th parallel line, we need to find the length of the 10th parallel line, which is NK. We can use the given ratios DM: MN = 2:1 to further simplify the equation.

Since NH = 6x + 6y + 4 and NH is the 4th parallel line, we know that NK = NH + DM + MN. Therefore:

NK = (6x + 6y + 4) + (2z) + (z)
= 6x + 6y + 3z + 4

Substituting the given value of KE:
KE = 2 cm = 2x

Therefore, x = 1 cm.

Now we can substitute the value of x into the equation for NK:

NK = 6(1) + 6y + 3z + 4
= 6 + 6y + 3z + 4
= 10 + 6y + 3z

Since we know that NH is the 4th parallel line, we can use the ratios:
NH = 3x + 3y + 4
= 3(1) + 3y + 4
= 3 + 3y + 4
= 7 + 3y

Since NH is the 4th parallel line, we can deduce that NH = NK - 6y - 3z.
Therefore:

7 + 3y = 10 + 6y + 3z

6y + 3z = 3

Solving these equations simultaneously will give us the values of y and z. Once we have y and z, we can substitute them back into the equation for NK:

NK = 10 + 6y + 3z

Note: Since we don't have information about the values of y and z, we are unable to give a specific numerical answer for the length of the 10th parallel line.