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stepan [7]
2 years ago
8

∆ ABC is similar to ∆DEF and their areas are respectively 64cm² and 121cm². If EF = 15.4cm then find BC.​

Mathematics
1 answer:
lyudmila [28]2 years ago
4 0

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ ∆ ABC is similar to ∆DEF

★ Area of triangle ABC = 64cm²

★ Area of triangle DEF = 121cm²

★ Side EF = 15.4 cm

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Side BC

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Since, ∆ ABC is similar to ∆DEF

[ Whenever two traingles are similar, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ]

\therefore \tt \boxed{  \tt \dfrac{area( \triangle \: ABC )}{area( \triangle \: DEF)} =  { \bigg(\frac{BC}{EF} \bigg)}^{2}   }

❍ <u>Putting the</u><u> values</u>, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

\implies  \tt  \dfrac{64   \: {cm}^{2} }{12 \:  {cm}^{2} }  =  { \bigg( \dfrac{BC}{15.4 \: cm} \bigg) }^{2}

❍ <u>By solving we get,</u>

\implies  \tt    \sqrt{\dfrac{{64 \: cm}^{2} }{ 121 \: {cm}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \sqrt{\dfrac{{(8 \: cm)}^{2} }{  {(11 \: cm)}^{2} }}   =   \bigg( \dfrac{BC}{15.4 \: cm} \bigg)

\implies  \tt    \dfrac{8 \: cm}{11 \: cm}    =   \dfrac{BC}{15.4 \: cm}

\implies  \tt    \dfrac{8  \: cm \times 15.4 \: cm}{11 \: cm}    =   BC

\implies  \tt    \dfrac{123.2 }{11 } cm   =   BC

\implies  \tt   \purple{  11.2 \:  cm}   =   BC

<u>Hence, BC = 11.2 cm.</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

\rule{280pt}{2pt}

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