Circumference 64104

The ABC triangle has a circumference of 11 cm. Triangle A'B'C ', similar to triangle ABC, has side lengths of 6 cm, 120 mm, and 1.5 dm larger than triangle ABC. Calculate the area of ​​the triangle A'B'C '.

Correct answer:

S₂ =  60.7947 cm²

Step-by-step explanation:

$$\begin{gathered} o_1=11\text{cm} \\ x=6\text{cm} \\ y=120\text{mm}\to \text{cm}=120:10\text{cm}=12\text{cm} \\ z=1.5\text{dm}\to \text{cm}=1.5\cdot 10\text{cm}=15\text{cm} \\ {o}_2=o_1+(x+y+z)=11+(6+12+15)=44\text{cm} \\ k=o_2/o_1=44/11=4 \\ X=x+a=k\cdot a \\ Y=y+b=k\cdot b \\ Z=z+c=k\cdot c \\ a=x/(k-1)=6/(4-1)=2\text{cm} \\ X=x+a=6+2=8\text{cm} \\ b=y/(k-1)=12/(4-1)=4\text{cm} \\ Y=y+b=12+4=16\text{cm} \\ c=z/(k-1)=15/(4-1)=5\text{cm} \\ Z=z+c=15+5=20\text{cm} \\ {s}_2=o_2/2=44/2=22\text{cm} \\ {S}_2=\sqrt{s_2\cdot (s_2-X)\cdot (s_2-Y)\cdot (s_2-Z)}=\sqrt{22\cdot (22-8)\cdot (22-16)\cdot (22-20)}=4\sqrt{231}\doteq 60.7947{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ {O}_1=a+b+c=2+4+5=11\text{cm} \\ {O}_2=X+Y+Z=8+16+20=44\text{cm} \end{gathered}$$

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