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3 years ago

Complete the similarity statement for the two given triangles.

Mathematics

1 answer:

tino4ka555 [31]3 years ago

4 0

The comment on top of me is a virus don’t go on it

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3 years ago

lyudmila [28]

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

★ ∆ 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. ]

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

❍ Putting the values, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

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

❍ By solving we get,

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

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

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

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

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

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

Hence, BC = 11.2 cm.

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

★ Figure in attachment.

$\rule{280pt}{2pt}$

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