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

Example 9 - Chapter 6 Class 10 Triangles - Part 2
Example 9 - Chapter 6 Class 10 Triangles - Part 3

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Example 9 In figure, the line segment XY is parallel to side AC of ABC and it divides the triangle into two parts of equal areas. Find the ratio / Given: ABC XY is parallel to AC i.e. XY II AC ar( BXY ) = ar(AXYC) To find : / Proof: In ABC & XBY, ABC = XBY ACB = XYB ABC ~ XBY ABC ~ XBY Now, we know that in similar triangles, Ratio of area of triangle is equal to ratio of square of corresponding sides ( )/( )=( / )^2 ( + )/( )=( / )^2 ( + )/( )=( / )^2 (2 )/( )=( / )^2 2=( / )^2 ( / )^2 = 2 / = 2 / 2= XB XB = / 2 Now, we need to find / Since, AX + XB = AB AX + / 2 = AB AX = AB / 2 AX = AB (1 1/ 2) / = ( 2 1)/ 2 Hence , / =( 2 1)/ 2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.