In the right angle triangle 4C 90°, AE and BD are temedians of a tria

1.	In the right angle triangle 4C 90°, AE and BD are temedians of a triangle ABC meeting at F. The ratio of thearea of 44BF and the quadrilateral FDCE is
Answer» Triangles BCD, BDA, and ABC has same height and their bases are related by CD = AD = AC/2.Hence, Area of BCD = Area of BDA = (Area of ABC)/2 Similarly, Area of ABE = Area of ACE = (Area of ABC)/2 So, Area of BCD = Area of ACE--> (Area of FDCE + Area of BEF) = (Area of FDCE + Area of AFD)--> Area of BEF = Area of AFD Now we know, Area of BCD = Area of BDA--> (Area of FDCE + Area of BEF) = (Area of AFB + Area of AFD)--> Area of FDCE = Area of AFB Hence, the required ratio is 1:1.

In the right angle triangle 4C 90°, AE and BD are temedians of a triangle ABC meeting at F. The ratio of thearea of 44BF and the quadrilateral FDCE is

Answer»

Triangles BCD, BDA, and ABC has same height and their bases are related by CD = AD = AC/2.Hence, Area of BCD = Area of BDA = (Area of ABC)/2

Similarly, Area of ABE = Area of ACE = (Area of ABC)/2

So, Area of BCD = Area of ACE--> (Area of FDCE + Area of BEF) = (Area of FDCE + Area of AFD)--> Area of BEF = Area of AFD

Now we know, Area of BCD = Area of BDA--> (Area of FDCE + Area of BEF) = (Area of AFB + Area of AFD)--> Area of FDCE = Area of AFB

Hence, the required ratio is 1:1.