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Question 13 - Area of Triangles - Areas of Parallelograms and Triangles

Last updated at May 29, 2023 by

Ex 9.3, 13 - ABCD is a trapezium with AB || DC. A line - Triangles with same base & same parallel lines

Ex 9.3, 13 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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Question 13 ABCD is a trapezium with AB ∥ DC . A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar(ACY). Given: A trapezium ABCD where AB ∥ DC & AC ∥ XY To prove: ar (ADX) = ar (ACY) Construction: Joining XC Proof : For ΔACX and ΔACY lie on the same base AC and are between parallel lines AC & XY ∴ ar(ACX) = ar(ACY) Also, for ΔACX and ΔADX lie on the same base AX and are between parallel lines AX & DC ∴ ar(ACX) = ar(ADX) From (1) & (2), ar( ACY ) = ar(ADX) Hence proved

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