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