Given: △CGF ,CG=GF=CF=and, KOPS is a square inside this triangle with one side on CF and its two vertices on CG and GF
Find: Area of KOPS

  AKPOS/ACGF = 28√3 -48 ≈ 0.497423

Step-by-step explanation:

Let's assume the square is a unit square. Then the height of the triangle is ...

  1 +(√3)/2 = h

and half the base of the triangle is ...

  h/√3 = (√3)/3 +1/2 = b/2

The area of the triangle is the product of these:

  A = (1/2)bh

  = (2 +√3)/2 × (3 +2√3)/6

  = (6 +3√3 +4√3 +6)/12 = (12 +7√3)/12

So, the ratio of the area of the square to that of the triangle is the inverse of this, or ...

  (square area)/(triangle area) = 12/(12+7√3)

  (square area)/(triangle area) = 28√3 -48

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.