Perimeter 66414

The perimeter of triangle ABC is 162 dm. The lengths of its sides are in the ratios a:b = 2:3 and a:c = 8:7. Determine the lengths of the sides of the triangle.

Correct answer:

a =  48 dm
b =  72 dm
c =  42 dm

Step-by-step explanation:

$$\begin{gathered} o=162\text{dm} \\ o=a+b+c \\ a=2/3\cdot b \\ a=8/7\cdot c \\ 162=a+b+c \\ a=2/3\cdot b \\ a=8/7\cdot c \\ a+b+c=162 \\ 3a-2b=0 \\ 7a-8c=0 \\ Pivot:Row1\leftrightarrow Row3 \\ 7a-8c=0 \\ 3a-2b=0 \\ a+b+c=162 \\ Row2-\frac{3}{7}\cdot Row1\to Row2 \\ 7a-8c=0 \\ -2b+3.429c=0 \\ a+b+c=162 \\ Row3-\frac{1}{7}\cdot Row1\to Row3 \\ 7a-8c=0 \\ -2b+3.429c=0 \\ b+2.143c=162 \\ Row3-\frac{1}{-2}\cdot Row2\to Row3 \\ 7a-8c=0 \\ -2b+3.429c=0 \\ 3.857c=162 \\ c=\frac{162}{3.85714286}=42 \\ b=\frac{0-3.4285714285714c}{-2}=\frac{0-3.42857143\cdot 42}{-2}=72 \\ a=\frac{0+8c}{7}=\frac{0+8\cdot 42}{7}=48 \\ a=48 \\ b=72 \\ c=42 \end{gathered}$$

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