1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
  3. Ex 9.3

Ex 9.3, 4

Last updated at Dec. 8, 2016 by

Ex 9.3, 4 - In figure, ABC and ABD are two triangles - Median divides triangle into two triangles of equal area

  1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
    3. Ex 9.3
Ask Download

Transcript

Ex 9.3, 4 In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD). Given: ΔABC and ΔABD on the same base AB & AB bisects CD, i.e. , OC = OD To prove: ar (ABC) = ar (ABD) Proof : In Δ ACD, Since OC = OD ∴ OA is the median. ⇒ ar(Δ AOC) = ar(Δ AOD) Similarly , in Δ BCD Since OC = OD ∴ OB is the median ⇒ ar(Δ BOC) = ar(Δ BOD) Adding (2) & (3) ar(Δ AOC) + ar(Δ BOC) = ar(Δ AOD) + ar(Δ BOD) ar(Δ ABC) = ar(Δ ADB) Hence proved

Examples →
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail