Geometria

D) ¿Existen segmentos paralelos en tu trazo? Si (AB-FE, BC-GF, CD-HG, DE-HA)

CONDICIÓN DE PARALELISMO:
m1=m2
PENDIENTES DE (AB, CD, EF y HG= 0)
m= y1 –y2
x1-x2
m=6-4
2-2
m=0
PENDIENTES DE (BC, FG = -0.9929)
m= 6-7.41
6.83-5.41
m= -1.41
1.42
m= -0.9929
PENDIENTES DE (DE, AH= 0.9929)
m= 2.59-4
5.41-6.83

m= -1.41
-1.42
m=0.9929

AB= EF
0=0
CD=HG
0=0
BC=GF
-0.9929=-0.9929
DE=AH
0.9929=0.9929

E) ECUACIONES DE LAS RECTAS DE CADA LADO
y- y1= y1-y2 (x-x1)x1-x2

AB:
y- 6 = 6-4 (x-2)
2-2
y- 6 =2(x-2)
y- 6 =2x-4
0 =2x+2-y
BC
y- 4 = 4-2.59 (x-2)
2-3.41
y- 4 = -1 (x-2)
y- 4 = -x+2
0 = x-6+y
CD
y- 2.59 =2.59-2.59 (x-3.41)
3.41-3.41
y- 2.59 = x-3.41
0 = x-0.82-y
DE
y- 2.59 = 2.59-4 (x-5.41)
5.41-6.38
y- 2.59 = -1.41 (x-5.41)
-1.42
y-2.59 = 0.99 (x-5.41)
y- 2.59 = 0.99x- 5.371
0 = 0.99x+7.96-y
EF
y- 4 = 4-6 (x-6.83)
6.83-6.83
y- 4 = -2 (x-6.83)
y- 4 =-2x+13.66
0 = -2x+17.66-y
FG
y- 6 =6-7.41 (x-6.83)
6.83-6.83
y- 6 = x-6.83
0 = x-0.83-y
AH
y- 6 = 6-7.41 (x-2)
2-3.41
y- 6 = x-2
0 = x+4-y


F) ECUACIONES DE LAS RECTAS DE LOS APOTEMAS

AI
y–6 = 6-5(x-2)
2-4.41
y- 6 = 1(x-2)
-2.41
y- 6 =0.414x-0.829
0 = 0.414x-5.171+y
BI
y- 4 = 4-5 (x-2)
2-4.41
y- 4 = -1 (x--2)
-2.41
y-4 = 0.414x-0.8298
0= 0.414x-3.1702-y
CI
y- 2.59 = 2.59-5 (x-3.41)
3.41-4.41

y- 2.59 = -2.41 (x-3.41)
-1
y- 2.59 = 2.41x-8.2181
0 =2.41x-5.62-y
DI
y- 2.59 = 2.59-5 (x-5.41)
5.41-4.41
y- 2.59 = -2.41 (x-5.41)
1
y- 2.59 = -2.41x-13.03
0 = -2.41x-1.448-y
EI
y- 4 = 4-5 (x...