If triangle DEF similar to triangle KLM such that DE=(2x+2)cm,DF=(5x-7)cm,KL=2cm,KM=3cm and EF=10cm,find the LM.

DE/KL = DF/KM = EF/LM

(2x+2)/2 = (5x-7)/3
6x+6 = 10x-14
x = 5

10/LM = x+1 = 6
LM = 5/3

Thank you

LM=5/3

LM=5/3

LM 5/3c.m

Why did the triangle go to the comedy club?

To catch a tangential line!

Now, let's solve your problem. Since triangle DEF is similar to triangle KLM, we can set up a proportion:

(DE / KL) = (EF / LM)

Replacing the given values:

((2x + 2) / 2) = (10 / LM)

Cross-multiplying:

(2x + 2) * LM = 2 * 10

Simplifying:

2x * LM + 2 * LM = 20

Dividing both sides by 2:

x * LM + LM = 10

Factoring out LM:

LM * (x + 1) = 10

Dividing both sides by (x + 1):

LM = 10 / (x + 1)

And there you have it! LM is equal to 10 divided by the sum of x and 1.

To find the length of segment LM, we first need to determine the value of x using the given information and solve for it.

Since triangle DEF is similar to triangle KLM, we can set up a proportion using the corresponding side lengths:

DE / KL = EF / LM

Substituting the given values:

(2x+2) / 2 = 10 / LM

To solve for x, we can cross-multiply:

10(2x+2) = 2 * LM

Simplifying the equation:

20x + 20 = 2 * LM

Next, we'll find the length of segment LM by plugging in the given values of KL = 2cm, KM = 3cm, and EF = 10cm.

Substituting these values into the equation, we have:

20x + 20 = 2 * 3

20x + 20 = 6

Subtracting 20 from both sides:

20x = 6 - 20

20x = -14

Dividing both sides by 20:

x = -14 / 20

Simplifying:

x = -7 / 10

Now we need to find the length of LM. Plugging in the value of x into one of the original given equations involving LM, let's use the proportion:

DE / KL = EF / LM

(2x+2) / 2 = 10 / LM

Substituting x = -7 / 10:

(2(-7/10) + 2) / 2 = 10 / LM

(-14/10 + 2) / 2 = 10 / LM

Simplifying the numerator:

(-14 + 20) / 10 = 10 / LM

6 / 10 = 10 / LM

Cross-multiplying:

6 * LM = 10 * 10

6 * LM = 100

Dividing both sides by 6:

LM = 100 / 6

Simplifying:

LM ≈ 16.67 cm

Therefore, the length of LM is approximately 16.67 centimeters.