**Q1)**As the ÐAOB = ÐCOD, length of arc AB = length of arc CD

Answer option “

**C**”

**Q2)**As the ÐAOC = ÐBOD, length of arc AC = length of arc BD

Answer option “

**C**”

**Q3)**Given that ÐAOB is acute --> (1)

We know that ÐBOD = 180

^{0}– ÐAOB --> (2)
From (1) and (2) we can conclude that ÐBOD measures obtuse angle

and hence, ÐBOD > ÐAOC

and hence, Length of arc BD > Length of arc AB.

Answer option is “

**B**”

**Q4)**As the ÐAOB = ÐCOD,

Area of shaded region AOB = Area of shaded region COD

Answer option “

**C**”

**Q5)**As the ÐAOC = ÐBOD,

Area of un-shaded region AOC = Area of un-shaded region BOD

Answer option “

**C**”

**Q6)**Given that ÐAOB = 60

^{0}

Radius of the circle = 4 units

Length of arc AB = (angle made by the arc at the centre)*(2πr)/360

^{0}
= 60 * 2 * π * 4/360 =

**4 π/3 units**

**Q7)**Given that radius of the circle = r = 3 units

ÐAOB = 60

^{0}
ÐCOD = ÐAOB = 60

^{0}
Total angle made in the shaded region = ÐAOB + ÐCOD

= 60

^{0}+ 60^{0}= 120^{0}
Total area of shaded region = (total angle made by the arc at the centre)*(πr

^{2})/360^{0}
= 120 * π * 3

^{2 }/ 360 =**3 π Sq.Units****Q8)**Given that, radius = r = 3 and ÐAOB = 60

^{0}

We know that AO = BO = r = 3

Consider ∆AOB, where in two sides AO and BO are equal in length

Hence angles opposite to these sides are equal,

ie., ÐOAB = ÐOBA

ÐOAB + ÐOBA + ÐAOB = 180

^{0}
ÐOAB + ÐOBA = 180

^{0 }- 60^{0}= 120^{0}
ÐOAB = ÐOBA = 120

^{0}/2 =60^{0}
As all the angles in ∆OAB are equal to 60

hence length of AB = length of OA = length of OB = r =^{0}, it is an equilateral triangle and**3 units**

**Q9)**Let ÐAOB = α

Length of arc AB = α * 2 * π * r/360 --> (1)

Angle made by arc ACDB at centre = 360- α

Length of arc ACDB = (360-α) * 2 * π * r/360 --> (2)

Given that, Length of arc ACDB = 2 * Length of arc AB --> (3)

From (1), (2) and (3),

(360-α) * 2 * π * r/360 = 2 * α * 2 * π * r/360

(360-α) = 2 α => 360 = 3α => α = **120**

^{0}
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