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

• 1st Method
• 2nd Method
• 3rd Method

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 = 2x

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

⇒ AE² = 4x² + 4x² + 2 × 2x × 2x × ½ = 12x²

then, AE = 2x√3

⇒ BD = AE = 2x√3

Area of ΔAFE = ½ × AF × FE × sin 120°

⇒ Area of ΔAFE = ½ × 2x × 2x × √3/2

so, Area of ΔAFE = x²√3

⇒ Area of ΔBCD = Area of ΔAFE = x²√3

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

⇒ Area of ΔPBD = Area of ΔAPE = x²√3

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

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

⇒ Red Area = 4x²√3

Area of PBCD = x²√3 + x²√3 = 2x²√3

⇒ Blue Area = 2x²√3

⇒ Red Area = 2 × Blue Area

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 = 2x

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

⇒ AE² = 4x² + 4x² + 2 × 2x × 2x × ½ = 12x²

⇒ AE = 2x√3

Area of ΔAFE = ½ × AF × FE × sin 120° = ½ × 2x × 2x × √3/2 = x²√3

Area of ΔAPE = ½ × AP × AE = ½ × x × 2x√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

Red Area = Area of Hexagon – Blue Area

Blue Area = Area of ΔPKD – Area of ΔBKC

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

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

⇒ Area of ΔPKD = 3x²√3

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

⇒ Area of ΔBKC = x²√3

Blue Area = 3x²√3 – x²√3 = 2x²√3

Area of Hexagon = 6 × ¼ × √3× AB² = 6 × ¼ × √3× 4x²

⇒ Area of Hexagon = 6x²√3

Red Area = 6x²√3 – 2x²√3 = 4x²√3

⇒ Red Area = 2 × Blue Area