Answer :

The figure for the above question is

So, arc AB = 44cm, radius, r = BO = AO = 42cm………(i)

We know the length of the arc can be written as

Substituting the values, we get

θ = 60°……(ii)

So, the angle of the minor segment is 60°.

The minor segment can be divided into two as shown in the figure.

So, we know the

area of segment ABX = area of sector AOBX – area of ΔAOB….(iii)

We know,

area of sector AOBX

Substituting the values from equation (i) and equation (ii), we get

area of sector AOBX

area of sector AOBX

area of sector AOBX = 22 × 7 × 6

area of sector AOBX = 924cm^{2}………..(iv)

Now consider ΔAOB, and draw a line AM perpendicular to AB, i.e.,

Here AO = OB,

∴ by RHS congruency ΔAMO≅ΔBMO

Now by CPCT (Corresponding parts of congruent triangles are equal), so AM = MB⇒ AB = 2AM………….(v)

Hence M is the mid - point of AB and

∠AOM = ∠BOM =

Now consider right - angled ΔAMO, we know

Now substituting the corresponding values, we get

But , substituting this value in above equation, we get

⇒ OM = 21√3cm…..(vi)

And also, in same triangle,

Now substituting the corresponding values, we get

But ,

substituting this value in above equation, we get

⇒ AM = 21cm….(vii)

Now the area of ΔABO

Substituting values from equation (v) and (vi), we get

Area of ΔABO

Substituting values from equation (vii), we get

Area of ΔABO

Cancelling the like terms, we get

Area of ΔABO = 441√3cm^{2}……(viii)

So, from equation (iii), we have

area of segment ABX = area of sector AOBX –area of ΔAOB

Substituting values from equation (iv) and (viii) in above equation, we get

area of segment ABX = (924 - 441√3) cm^{2}

Hence the area of the minor segment of the given circle is (924 - 441√3)cm^{2}

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