In each picture below, the explanation of the green part is given. Calculate in each, the probability of a dot put without looking to be within the green part.
A triangle got by joining alternate vertices of a regular hexagon.

The figure is shown below:

Let the side of this hexagon be “x”.
Then, in this hexagon, a total of 6 equilateral triangles can be made, which are:
ΔOED, ΔODC, ΔOCB, ΔOBA, ΔOAF, ΔOFE.
Now, in the figure given below:

In ΔOPF and ΔOPB,
∠POF = ∠POB (angle in equilateral triangle = 600)
OF = OB (sides of equilateral triangle)
OP is common in both the triangles.
Hence, we can say that ΔOPF ≅ ΔOPB
⇒ ![]()
Similarly, we can prove that ΔAPF ≅ Δ APB
⇒ AP = PO = ![]()
Now in, right ΔOPF,
OP2 + PF2 = OF2
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒ b = √3x
Now,
Area of hexagon with side “x” = ![]()
Area of big triangle (green shaded) = ![]()
![]()
![]()
![]()
Probability of dot is putted in the green part ![]()

![]()
= 0.083
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.


