Solve all your doubts with Teachoo Black (new monthly pack available now!)

Are you in **school**? Do you **love Teachoo?**

We would love to talk to you! Please fill this form so that we can contact you

Examples

Example 1

Example 2 Important

Example 3

Example 4

Example 5 Important

Example 6

Example 7 Important

Example 8 Important

Example 9 Important Deleted for CBSE Board 2023 Exams You are here

Example 10 Important Deleted for CBSE Board 2023 Exams

Example 11 Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Chapter 6 Class 10 Triangles

Serial order wise

Last updated at Aug. 4, 2021 by Teachoo

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