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
1 answer:
Answer:
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
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