# Relation between two Areas Inside a Regular Hexagon

## Geometry Math Problem: Find the relation between red area and blue area inside a regular hexagon

The figure shows a regular hexagon ABCDEF, P is the mid-point of AB. Hexagon is divided into two geometry shapes through PD. then find the relation between Red Area and Blue Area

## Solution

We can solve this problem in different ways, three of them are shown here

## Method 1 : Using cosine formula and Area of triangles

Red Area = Area of APDEF

Blue Area = Area of PBCD

Area of APDEF = Area of ΔAFE + Area of ΔAPE + Area of ΔPED

Area of PBCD = Area of ΔBCD + Area of ΔPBD

Let AP = BP =* x* then, AF = FE = 2*x*

AE² = AF² + FE² − 2 × AF × FE × cos 120°

⇒ AE² = 4*x*² + 4*x*² + 2 × 2*x* × 2*x* × *½* = 12*x*²

then, AE = 2*x√*3

⇒ **BD = AE = 2 x√3**

Area of ΔAFE = *½* × AF × FE × sin 120°

⇒ Area of ΔAFE = *½* × 2*x* × 2*x* × *√*3/2

so, Area of ΔAFE = *x*²*√*3

⇒ **Area of ΔBCD = Area of ΔAFE = x²√3**

Area of ΔAPE = *½* × AP × AE = *½* × *x* × 2*x√*3 = *x²√*3

⇒ **Area of ΔPBD = Area of ΔAPE = x²√3**

**Area of ΔAPE** = *½* × ED × AE = *½* × 2*x* × 2*x√*3 =** 2 x²√3**

Area of APDEF** **=* x²√*3 + 2*x²√*3 + *x²√*3 =** **4*x²√*3

⇒ **Red Area = 4 x²√3**

**Area of PBCD **= *x²√3 + x²√3 = 2x²√3*

⇒ **Blue Area = 2 x²√3**

⇒ **Red Area = 2 ×** **Blue Area**

## Method 2 : Using cosine formula and Equal Area triangles

In this method, we divide rectangle ABDE into Four equal triangles. hence we get a total of six triangles. The area of the four triangles is equal to each other, also the area of the other two triangles is equal to each other.

Let AP = BP =* x* then, AF = FE = 2*x*

AE² = AF² + FE² − 2 × AF × FE × cos 120°

⇒ AE² = 4*x*² + 4*x*² + 2 × 2*x* × 2*x* × *½* = 12*x*²

⇒ AE = 2*x√*3

Area of ΔAFE = *½* × AF × FE × sin 120° = *½* × 2*x* × 2*x* × *√*3/2 = *x*²*√*3

Area of ΔAPE = *½* × AP × AE = *½* × *x* × 2*x√*3 = *x²√*3

⇒ Area of ΔAFE = Area of ΔAPE so all six triangle has same area

Red Area = Area of APDEF = 4 × *x²√*3

Blue Area = Area of PBCD = 2 × *x²√*3

⇒ **Red Area = 2 ×** **Blue Area**

## Method 3 : Using Area of Hexagon

Red Area = Area of Hexagon – Blue Area

Blue Area = Area of ΔPKD – Area of ΔBKC

Let AP = BP =* x* then, BK = CK = CD = 2*x*

Area of ΔPKD = *½* × PK × DK × sin 60° = *½* × 3*x* × 4*x* × *√*3/2

⇒ **Area of ΔPKD = 3 x²√3**

Area of ΔBKC = *½* × BK × CK × sin 60° = *½* × 2*x* × 2*x* × *√*3/2

⇒ **Area of ΔBKC = x²√3**

**Blue Area **= 3*x ²√*3 –

*x*3 =

*²*√**2**

*x*3*²*√Area of Hexagon = 6 × *¼ × √*3× AB* ²* = 6 ×

*¼ × √*3× 4

*x*

*²*⇒ Area of Hexagon = 6* x²√*3

**Red Area **= 6* x²√*3 –

*2*=

*x*3*²*√**4**

*x*3*²*√⇒ **Red Area = 2 ×** **Blue Area**