Two similar

Two similar triangles, one has a circumference of 100 cm, the second has sides successively 8 cm, 14 cm, and 18 cm longer than the first. Find the lengths of its sides.

Correct answer:

a1 =  20 cm
b1 =  35 cm
c1 =  45 cm
a2 =  28 cm
b2 =  49 cm
c2 =  63 cm

Step-by-step explanation:

o1=100 cm o1 = a1+b1+c1 o2 = a2+b2+c2 a2 = a1+8 b2 = b1+14 c2 = c1+18  o2 = k  o1 a1+8+b1+14+c1+18 = k   o1  o1 + 8+14+18 = k   o1 k=o1o1+8+14+18=100100+8+14+18=57=1.4  o2=k o1=57 100=57 100=5700=140 cm  a2 = a1+8 k a1 = a1+8  a1=k18=5718=20 cm
b1=k114=57114=35 cm
c1=k118=57118=45 cm
a2=k a1=57 20=57 20=5140=28 cm
b2=k b1=57 35=57 35=5245=49 cm
c2=k c1=57 45=57 45=5315=63=63 cm   Verifying Solution:  O1=a1+b1+c1=20+35+45=100 cm O2=a2+b2+c2=28+49+63=140 cm



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