Two similar

Two similar triangles, one has a circumference of 100 cm, the second has sides successively 8 cm, 14 cm, 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=o1+8+14+18o1=100+8+14+18100=75=1.4  o2=k o1=1.4 100=140 cm  a2=a1+8 k a1=a1+8  a1=8k1=81.41=20 cm
b1=14k1=141.41=35 cm
c1=18k1=181.41=45 cm
a2=k a1=1.4 20=28 cm
b2=k b1=1.4 35=49 cm
c2=k c1=1.4 45=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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