The chord AB of a circle with centre at O is 2√3 times the height of the minor segment. If P is the area of the sector OAB and Q is the area of the minor segment of the circle, then what is the approximate value of P/Q?

(Take √3= 1.7 and π = 3.14)

Correct option is B

Given:

The chord AB of a circle with center at O is 2√3√3 times the height of the minor segment.

P is the area of the sector OAB and Q is the area of the minor segment of the circle.

Solution:

Let the radius of circle = r

AC = BC = √3

OC = r - 1

Using Pythagoras theorem on ΔOCA,

r² = (√3)²+ (r -1)²

r2 = 3 + r² + 1 - 2r

r = 2

OC = 2 -1 = 1

Ratio of sides = 2 : 2 : 2√3 = 1 : 1 : √3

If ratio of sides of a triangle is 1 : 1 : √3 than the angle opposite to side √3 will be 120°.

P = Area of sector OAB

Q = Area of minor segment under AB

P = π × 2² × (120/360)

Q = [π × 22 × (120/360)] - Area of triangle

Q =  [π × 22 × (120/360)] - (2 × 2 × sin 120°)/2

Q = [π × 22 × (120/360)] - (2 × 2 × (√3/2))/2

Q = (π × 4)/3 - √3

P/Q = $$\frac{\cfrac{4 \pi}{3}}{\cfrac{4 \pi}{3} - \sqrt{3}}$$

P/Q = 1.7

∴ The correct answer is 1.7.

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