#### To determine

a)

**To find:**

The area of polygon

#### Answer

*nr22sin2πn*

#### Explanation

**1) Formula:**

i.

*Area of triangle=12bc·sinθ*

*w*here b & c are *the *sides of *the *triangle and θ is *the *angle between b & c

ii.

*limn→0sinθθ=1*

2) **Calculation:**

It is given* that An* *is* the area of a polygon with n equal sides inscribed in circle with radius r.

*The p*olygon dividing into n congruent triangles with central angle 2πn.

In each of these triangles, *the *length of two sides is r, and the angle between them is θ=2πn.

Therefore,

*Area of triangle=12bc·sinθ*

*=12·r·r·sin2πn*

*=r22·sin2πn*

So,Area of triangle=r22·sin2πn

Area of the whole polygon = n times the area of triangle

*=n·r22·sin2πn*

*=nr22sin2πn*

**Conclusion:**

The area of the whole polygon =nr22sin2πn

#### To determine

b)

**To show:**

*limn→0 An=πr2*

#### Answer

*limn→0 An=πr2*

#### Explanation

1) **Calculation:**

From part a),

*An=nr22sin2πn*

Take limit as n→∞ on both sides.

*limn→∞ An=limn→∞nr22sin2πn*

*=r22limn→∞nsin2πn*

Multiply and divide by 2πn on the right side of the equation.

*=r22limn→∞nsin2πn2πn·2πn*

Simplify.

*=r22·2πlimn→∞sin2πn2πn*

Now substitute x=1n. Thus, the limn→∞ becomes limx→0 since x→0 as n→∞

*limx→0 An=r22·2π limx→0sin2πx2πx*

By using limit formula,

*=r2×π1*

*limx→0 An=πr2*

Now substitute n back into the equation.

*limn→∞An=πr2*

**Conclusion:**

*limn→0An=πr2*