# Triangles ABC and DEF are similar triangles. Use this fact

Triangles ABC and DEF are similar triangles. Use this fact to solve the exercise. Round to the nearest tenth.
Find the perimeter of triangle ABC.

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Befory

Concept used:
(1) Pythagoras theorem: In a right angle triangle
$$\displaystyle{\left({H}{y}{p}{o}{t}{e}nu{s}{e}\right)}^{{{2}}}={\left({P}{e}{r}{p}{e}{n}{d}{i}{c}ul{{a}}{r}\right)}^{{{2}}}+{\left({B}{a}{s}{e}\right)}^{{{2}}}$$
(2) If two triangles are similar, their perimeter have the exact same ratio.
In right angle triangle DEF
$$\displaystyle{\left({H}{y}{p}{o}{t}{e}nu{s}{e}\right)}^{{{2}}}={\left({P}{e}{r}{p}{e}{n}{d}{i}{c}ul{{a}}{r}\right)}^{{{2}}}+{\left({B}{a}{s}{e}\right)}^{{{2}}}$$
$$\displaystyle{\left({D}{E}\right)}^{{{2}}}={\left({E}{F}\right)}^{{{2}}}+{\left({F}{D}\right)}^{{{2}}}$$
$$\displaystyle{\left({D}{E}\right)}={\left({15}\right)}^{{{2}}}+{\left({6}\right)}^{{{2}}}$$
$$\displaystyle={225}+{1296}$$
$$\displaystyle={1521}$$
$$\displaystyle{D}{E}=\pm\sqrt{{{1521}}}$$
$$\displaystyle=\pm{139}$$
Side can't be negative so DE =39
By similar triangle property:
$$\displaystyle{\frac{{{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{D}{E}{F}}}{{{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{A}{B}{C}}}}={\frac{{{D}{F}}}{{{A}{C}}}}={\frac{{{D}{E}}}{{{A}{B}}}}={\frac{{{E}{F}}}{{{B}{C}}}}$$

$$\displaystyle{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{D}{E}{F}={39}+{36}+{15}={90}$$

$$\displaystyle{\frac{{{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{D}{E}{F}}}{{{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{A}{B}{C}}}}={\frac{{{D}{F}}}{{{A}{C}}}}$$

$$\displaystyle{\frac{{{90}}}{{{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{A}{B}{C}}}}={\frac{{{36}}}{{{12}}}}$$

$$\displaystyle{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{A}{B}{C}={\frac{{{90}\times{12}}}{{{36}}}}$$

$$\displaystyle{p}{e}{r}{i}{m}{e}{t}{e}{r}\ {o}{f}\ \triangle{A}{B}{C}={30}$$