BDEF is a square formed inside isolescale right angled at B show that area of triangle. AEF =area of triangle CDE
Answers
Answered by
5
Hey there,
by making a diagram from the given information we will get
AB = BC
AB - BF = BC - BD
and AF = CD say 1
BD = DE = EF = BF say 2
so the area of ΔAEF = 1/2 x AF x EF
area of triangle CDE = 1/2 CD x DE
or 1/2 AF x EF from equation 1 and 2
so the area of ΔAEF = area of Δ CDE
Hope this helps!
by making a diagram from the given information we will get
AB = BC
AB - BF = BC - BD
and AF = CD say 1
BD = DE = EF = BF say 2
so the area of ΔAEF = 1/2 x AF x EF
area of triangle CDE = 1/2 CD x DE
or 1/2 AF x EF from equation 1 and 2
so the area of ΔAEF = area of Δ CDE
Hope this helps!
suru18pallu:
But can you draw its diagram too
https://s3mn.mnimgs.com/img/shared/content_ck_images/images/AAAAA(4).png
Answered by
0
bro it very simple
just first try to construct your diagram and from the diagram u can see
FE=ED( Sides of a square are of same side)
now it is tricky one
AB-BF=BC-BF
WHY??? AS BECAUSE SIDES OF A SQUARE R SAME
so therefore BF=FE=ED=BD are same
so now compare the area of the triangle
ar(AFE)=1/2×AF×EF
-----------------------------
ar(CDE)=1/2×CD×ED
as it was proved earlier that the sides are same
so therefore
exchange the place CD×ED by AF×EF
SO ALL WILL BE CUTED OUY AND BRING THE ar(CDE) TO THE L.H.S OF THE EQUAL
then it will be proved
ar(AFE)=ar(CDE)
just first try to construct your diagram and from the diagram u can see
FE=ED( Sides of a square are of same side)
now it is tricky one
AB-BF=BC-BF
WHY??? AS BECAUSE SIDES OF A SQUARE R SAME
so therefore BF=FE=ED=BD are same
so now compare the area of the triangle
ar(AFE)=1/2×AF×EF
-----------------------------
ar(CDE)=1/2×CD×ED
as it was proved earlier that the sides are same
so therefore
exchange the place CD×ED by AF×EF
SO ALL WILL BE CUTED OUY AND BRING THE ar(CDE) TO THE L.H.S OF THE EQUAL
then it will be proved
ar(AFE)=ar(CDE)
Similar questions