Circumferences 83111

Péta composed several planar shapes from mutually congruent triangles. The circumferences of the first three are 8 cm, 11.4 cm, and 14.7 cm, respectively. Determine the perimeter of the fourth shape.

Correct answer:

o₄ =  10 cm

Step-by-step explanation:

$$\begin{gathered} \Delta o=a+b+c \\ {o}_1=8\text{cm} \\ {o}_2=11.4\text{cm} \\ {o}_3=14.7\text{cm} \\ {o}_1=2a+2b \\ {o}_2=2a+2c \\ {o}_3=3\cdot (a+b+c)-(a+b+c)=2\cdot (a+b+c)=2a+2b+2c \\ {o}_1=2a+2b \\ {o}_2=2a+2c \\ {o}_3=2a+2b+2c \\ 8=2a+2b \\ 11.4=2a+2c \\ 14.7=2a+2b+2c \\ 8=2\cdot a+2\cdot b \\ 11.4=2\cdot a+2\cdot c \\ 14.7=2\cdot a+2\cdot b+2\cdot c \\ 2a+2b=8 \\ 2a+2c=11.4 \\ 2a+2b+2c=14.7 \\ Row2-Row1\to Row2 \\ 2a+2b=8 \\ -2b+2c=3.4 \\ 2a+2b+2c=14.7 \\ Row3-Row1\to Row3 \\ 2a+2b=8 \\ -2b+2c=3.4 \\ 2c=6.7 \\ c=\frac{6.7}{2}=3.35 \\ b=\frac{3.4-2c}{-2}=\frac{3.4-2\cdot 3.35}{-2}=1.65 \\ a=\frac{8-2b}{2}=\frac{8-2\cdot 1.65}{2}=2.35 \\ a=\frac{47}{20}=2.35 \\ b=\frac{33}{20}=1.65 \\ c=\frac{67}{20}=3.35 \\ {o}_4=2\cdot b+2\cdot c=2\cdot 1.65+2\cdot 3.35=10\text{cm} \end{gathered}$$

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