# I do not know the right rules to use in order to solve this problem. It seems simple, but I think I'm a little tired so I'm forgetting stuff. How do I solve this? Thanks!

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The main thing we have to know is that the length of a midline

(the line connecting midpoints of two sides) is the half of the length of the third side.

In this case, MN = (1/2)BC, ML = (1/2)AC, NL = (1/2)AB.

The perimeter in question is BM + MN + NC + BC. We know MN=8, find the rest.

Because BM = AM = (1/2)AB we obtain BM = NL = 6.

Because AN = CN = (1/2)AC we obtain CN = ML = 5.

And BC = 2*MN = 16.

So P = BM + MN + NC + BC = 6 + 8 + 5 + 16 = **35**.

The right answer is **(1) 35**

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Ohh, I see. I did not realize that ML and NL were midsegments as well. I really need to get my orientation fixed, I always mess up with that stuff! Thanks!

Given

In Triangle ABC, L,M,N are the mid points of BC, AB, AC. LM= 5, MN =8, NL = 6

**To find**

the perimeter of ABC

sol:

**From the Given link below the : see question 5**

`Delta ABC ~~ Delta AMN `

` Delta ABC ~~ Delta BML `

` Delta ABC ~~ Delta LNC`

as all these are similar triangles, now the corresponding ratios of the sides can be taken as follows

from the figure we can clearly see that `AB= 2AM , AC=2CN , BC = 2 BL`

so , from the first similarity we get `(AB)/(BC) = (AM)/(MN)`

`=> (2AM)/(BC) =(AM)/(MN) `

`=> BC = 2MN `

` => BC = 2 *8 =16`

from the SECOND similarity we get `(BC)/(AC) = (BL)/(ML)`

`=> (2BL)/(AC) =(BL)/(ML) `

`=> AC = 2ML `

`=> AC = 2*5 =10`

from the THIRD similarity we get `(AC)/(BA) = (CN)/(LN)`

`=> (2CN)/(BA) =(CN)/(LN) `

`=> BA = 2LN `

` => BA = 2*6 =12`

SO, The perimeter of the TRAPEZOID BMNC is = BM+MN+NC+BC = 6+8+5+16 = **35**

**Sources:**