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:

$$\begin{gathered} o_1=100\text{cm} \\ {o}_1=a_1+b_1+c_1 \\ {o}_2=a_2+b_2+c_2 \\ {a}_2=a_1+8 \\ {b}_2=b_1+14 \\ {c}_2=c_1+18 \\ {o}_2=k\cdot o_1 \\ {a}_1+8+b_1+14+c_1+18=k\cdot o_1 \\ {o}_1+8+14+18=k\cdot o_1 \\ k=\frac{o_1+8+14+18}{o_1}=\frac{100+8+14+18}{100}=\frac{7}{5}=1.4 \\ {o}_2=k\cdot o_1=\frac{7}{5}\cdot 100=\frac{7\cdot 100}{5}=\frac{700}{5}=140\text{cm} \\ {a}_2=a_1+8 \\ k\cdot a_1=a_1+8 \\ {a}_1=\frac{8}{k-1}=\frac{8}{\frac{7}{5}-1}=20\text{cm} \end{gathered}$$

$b_1=\frac{14}{k-1}=\frac{14}{\frac{7}{5}-1}=35\text{cm}$

$c_1=\frac{18}{k-1}=\frac{18}{\frac{7}{5}-1}=45\text{cm}$

$a_2=k\cdot a_1=\frac{7}{5}\cdot 20=\frac{7\cdot 20}{5}=\frac{140}{5}=28\text{cm}$

$b_2=k\cdot b_1=\frac{7}{5}\cdot 35=\frac{7\cdot 35}{5}=\frac{245}{5}=49\text{cm}$

$$\begin{gathered} c_2=k\cdot c_1=\frac{7}{5}\cdot 45=\frac{7\cdot 45}{5}=\frac{315}{5}=63=63\text{cm} \\ \text{ Verifying Solution: } \\ {O}_1=a_1+b_1+c_1=20+35+45=100\text{cm} \\ {O}_2=a_2+b_2+c_2=28+49+63=140\text{cm} \end{gathered}$$

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