Area

[Aman verma]

Q: In a right angled triangle, angle C = 90 degrees. AE and BD are two medians of the triangle ABC meeting at F. The ratio of the area of the triangle ABF and the quadrilateral FDCE is :

a) 1:1

b) 1:2

c) 2:1

d) 2:3

e) 1:3

[Anurag@Gurome]

Q: In a right angled triangle, angle C = 90 degrees. AE and BD are two medians of the triangle ABC meeting at F. The ratio of the area of the triangle ABF and the quadrilateral FDCE is :

Refer to the following figure,


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.

The correct answer is A.

[Aman verma]

Many thanks to Anurag sir.OA[spoiler]a)[/spoiler]. Regret unable to insert a diagram, don't know how to use the diagram feature. Anybody could guide, please do!