Question

# Use similarity in figures to solve for the value of the x

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Use similarity in figures to solve for the value of the x

2021-08-10

Here $$\displaystyle{D}{E}\parallel{A}{C}$$
From the figure
$$\displaystyle{\frac{{{B}{E}}}{{{B}{C}}}}={\frac{{{D}{E}}}{{{A}{C}}}}$$
$$\displaystyle\Rightarrow{\frac{{{6}}}{{{6}+{x}}}}={\frac{{{x}-{1}}}{{{2}{x}+{4}}}}$$
$$\displaystyle\Rightarrow{6}{\left({2}{x}+{4}\right)}={\left({6}+{x}\right)}{\left({x}-{1}\right)}$$
$$\displaystyle\Rightarrow{12}{x}+{24}={6}{x}-{6}+{x}^{{{2}}}-{x}$$
$$\displaystyle\Rightarrow{12}{x}+{24}={5}{x}+{x}^{{{2}}}-{6}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}+{5}{x}-{6}-{12}{x}-{24}={0}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}-{7}{x}-{30}={0}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}-{10}{x}+{3}{x}-{30}={0}$$
$$\displaystyle\Rightarrow{x}{\left({x}-{10}\right)}+{3}{\left({x}-{10}\right)}={0}$$
$$\displaystyle\Rightarrow{\left({x}-{10}\right)}{\left({x}+{3}\right)}={0}$$
$$\displaystyle\Rightarrow$$ either x-10=0 or, x+3=0
$$\displaystyle\Rightarrow{x}={10}\Rightarrow{x}=-{3}$$
x=-3 is not possitive since length is non negative
Hence x=10