#### notes

**(i) Pentagon :**

By constructing two diagonals AC and AD the pentagon ABCDE is divided into three parts.

So, area ABCDE = area of Triangle ABC + area of Triangle ACD + area of Triangle AED.

By constructing one diagonal AD and two perpendiculars BF and CG on it, pentagon ABCDE is divided into four parts.

So, area of ABCDE

= area of right angled D AFB + area of trapezium BFGC + area of right angled Triangle CGD + area of Triangle AED.

**(ii) Hexagon :**

**⦁ Method 1 : **

Since, it is a hexagon so NQ divides the hexagon into two congruent trapeziums.

Now area of trapezium MNQR = `2 xx {1/2 ("sum of parallel sides")` x `("perpendicular distance between them")}`

**⦁ Method 2 :**

Triangle MNO and Triangle RPQ are congruent triangles with altitude.

Area of hexagon MNOPQR= Area of Triangle MNO + Area of Triangle RPQ + Area of Rectangle MOPR

**(iii) Octagon:**Area of Octagonal Surface= Area of trapezium ABCH+ Area of rectangle HCDG + Area of Trapezium GDEF.