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Example 9 - Line segment XY is parallel to side AC of ABC - Area of similar triangles

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Example 9 In figure, the line segment XY is parallel to side AC of ฮ” ABC and it divides the triangle into two parts of equal areas. Find the ratio ๐ด๐‘‹/๐ด๐ต Given: ฮ” ABC XY is parallel to AC i.e. XY II AC ar(ฮ” BXY ) = ar(AXYC) To find : ๐ด๐‘‹/๐ด๐ต Proof: In ฮ” ABC & ฮ”XBY, โˆ  ABC = โˆ  XBY โˆ  ACB = โˆ  XYB ฮ” ABC ~ ฮ” XBY ฮ” ABC ~ ฮ” XBY Now, we know that in similar triangles, Ratio of area of triangle is equal to ratio of square of corresponding sides (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐ด๐ต๐ถ)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ)=( ๐ด๐ต/๐‘‹๐ต )^2 (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ + ๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ ๐ด๐‘‹๐‘Œ๐ถ)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ)=( ๐ด๐ต/๐‘‹๐ต )^2 (๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ + ๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ)/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ)=( ๐ด๐ต/๐‘‹๐ต )^2 (2 ร— ๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ )/(๐ด๐‘Ÿ๐‘’๐‘Ž ๐‘œ๐‘“ โˆ† ๐‘‹๐ต๐‘Œ)=( ๐ด๐ต/๐‘‹๐ต )^2 2=(๐ด๐ต/๐‘‹๐ต)^2 (๐ด๐ต/๐‘‹๐ต)^2 = 2 ๐ด๐ต/๐‘‹๐ต= โˆš2 ๐ด๐ต/โˆš2= XB XB = ๐ด๐ต/โˆš2 Now, we need to find ๐ด๐‘‹/๐ด๐ต Since, AX + XB = AB AX + ๐ด๐ต/โˆš2 = AB AX = AB โ€“ ๐ด๐ต/โˆš2 AX = AB (1 โ€“ 1/โˆš2) ๐ด๐‘‹/๐ด๐ต = (โˆš2 โˆ’ 1)/โˆš2 Hence , ๐ด๐‘‹/๐ด๐ต=(โˆš2 โˆ’1)/โˆš2

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