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Ex 6.4, 8 - ABC and BDE are two equilateral triangles - Area of similar triangles

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.4, 8 (Introduction) Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is • 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 Two equilateral triangle are always similar In ∆ 𝐴𝐵𝐶 𝑎𝑛𝑑 ∆ 𝐷𝐸𝐹 𝐷𝐸﷮𝐴𝐵﷯= 12﷮6﷯ 𝐸𝐹﷮𝐵𝐶﷯= 12﷮6﷯ 𝐷𝐹﷮𝐴𝐶﷯= 12﷮6﷯ Hence by SSS similarity ∆ 𝐴𝐵𝐶 ~ ∆ 𝐷𝐸𝐹 Ex 6.4, 8 Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 Given: ∆ 𝐴𝐵𝐶 is equilateral ∆ 𝐵𝐷𝐸 is equilateral BD = 1/2 𝐵𝐶 as D is midpoint of BC To find: (𝑎𝑟 ∆ 𝐴𝐵𝐶)/(𝑎𝑟 ∆ 𝐵𝐷𝐸) Solution: Since , ∆ 𝐴𝐵𝐶 and ∆ 𝐵𝐷𝐸 are equilateral, Their sides would be in the same ratio 𝐴𝐵/𝐵𝐸=𝐴𝐶/𝐸𝐷=𝐵𝐶/𝐵𝐷 Hence, by SSS similarity ∆ 𝐴𝐵𝐶 ~ ∆ 𝐵𝐷𝐸 And , we know that ratio of area of triangle is equal To the ratio of square of corresponding sides So, (𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐵𝐷𝐸)=(𝐵𝐶)^2/(𝐵𝐷)2 =(𝐵𝐶)^2/(𝐵𝐶/2)^2 =𝐵𝐶2/((𝐵𝐶^2)/4) =4𝐵𝐶2/𝐵𝐶2 = 4/1 Hence, (𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐵𝐷𝐸 )=4/1 i.e. 4 : 1 Option (C) is correct

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