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△ABC ~ △DEF. If area of △ABC = 9 sq. cm, area of △DEF = 16 sq. cm and BC = 2.1 cm, find the length of EF.

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Answer

Let the length of EF be x cm.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

∴Area of △ABCArea of △DEF=BC2EF2⇒916=(2.1)2x2⇒916=2.1×2.1x2⇒916=4.41x2⇒x2=4.41×169⇒x2=70.569⇒x2=7.84⇒x=7.84⇒x=2.8\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △DEF}} = \dfrac{BC^2}{EF^2} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{(2.1)^2}{x^2} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{2.1 \times 2.1}{x^2} \\[1em] \Rightarrow \dfrac{9}{16} = \dfrac{4.41}{x^2} \\[1em] \Rightarrow x^2 = \dfrac{4.41 \times 16}{9} \\[1em] \Rightarrow x^2 = \dfrac{70.56}{9} \\[1em] \Rightarrow x^2 = 7.84 \\[1em] \Rightarrow x = \sqrt{7.84} \\[1em] \Rightarrow x = 2.8

Hence, the length of EF = 2.8 cm.

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