Answers:
(a) BC = 40
(b) GF = 15
(c) CD = 45
(d) KM = 37.5
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Explanations:
Part (a)
GF is a midsegment of triangle ABC, so GF is half that of the parallel base AC
AC = 30 so GF = (1/2)*AC = 0.5*30 = 15
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Part (b)
For similar reasons as part (a), FB if half that of BC. This leads to FB = FC
FB = FC
FC = 20
since FB = 20
Now use the segment addition postulate
BC = BF + FC
BC = 20 + 20
BC = 40
Note: FB is the same as BF. The order of the letters does not matter.
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Part (c)
GF = AD are the same length because of the single tickmark
GF = 15 so AD = 15
use the segment addition postulate
CD = CA + AD
CD = 30 + 15
CD = 45
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Part (d)
EG = 15 since GF = 15 (and EG = GF by the single tickmark)
use the segment addition postulate
EF = EG + GF
EF = 15 + 15
EF = 30
The length of KM is the average of the base lengths EF and DC, since KM is a midsegment of the trapezoid
KM = (EF+DC)/2
KM = (30+45)/2
KM = (75)/2
KM = 37.5
