Triangles ABC and DEF are similar triangles. Use this fact

ostric16 2021-11-28 Answered
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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Expert Answer

Befory
Answered 2021-11-29 Author has 5987 answers

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


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


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


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


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


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

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