ABCD is a square of side length 1. EFGH are the points on its boundary such that AE=EB, 2BF=FC, 3CG=GD & 4DH=HA...

Question

ABCD is a square of side length 1.

If EFGH are the points on its boundary such that AE=EB, 2BF=FC, 3CG=GD & 4DH=HA then what is the area of the quadrilateral EFGH?

Answer 1

Assuming unit of length to be meter...
Area of ABCD = 1×1 = 1 m^2
Area of EFGH = Area of ABCD - Area of (AEH+EBF+FCG+GDH)

Area of AEH= 1/2*AE*AH= 1/2*1/2*4/5= 1/5
Area of EBF= 1/2*EB*BF= 1/2*1/2*1/3= 1/12
Area of FCG= 1/2*FC*CG= 1/2*2/3*1/4= 1/12
Area of GDH= 1/2*GD*DH= 1/2*3/4*1/5= 3/40

Area of (AEH+EBF+FCG+GDH)= 1/5+1/12+1/12+3/40= 53/120

So, Area of EFGH= 1 - 53/120= 67/120 m^2

Note: If we devide quadrilateral EFGH into 2 triangles and use Heron's method; A=square root (s(s-a)(s-b)(s-c)), where s=(a+b+c)/2.... we get the same result.

Answer 2

Area of ABCD = 1×1 = 1

Area of EFGH = Area ABCD - Area (EBF+FGC + GHD + HEA)

Area EFB = 1/2(1/2)(1/3) = 1/12
Area FGC = 1/2(2/3)(1/4) = 1/12
Area GHD = 1/2(3/4)(1/5) = 3/40
Area HEA = 1/2(4/5)(1/2) = 1/5

EBF + FGC + GHD + HEA = 1/12 +1/12 + 3/40 + 1/5
= 84/120 = 7/10

So,
Area EFGH = 1 - 7/10
= 3/10 length squared

Comments

I think you miscalculated a bit Justine...

Answer 3

66/120 square metres
As first find all the area of small triangles and the subtract it from are of square
1_53/120
67/120 m square

Answer 4

area of quadrilateral EFGH is 21/40

Answer 5

Size of highlighted area is =0.559
To be more specific ( I calculated areas of HAE=0.2 , HDG=0.075, GCF=0.083 FBE=0.083) sum of these areas = 0.441 subtracting that from whole area which is 1 answer 0.559