**Area of compound figures :**

You can find the areas of polygons by breaking the polygons into smaller shapes. Then we can apply area formulas you already know.

Let us see an example problems to understand this method

**Example 1 :**

Find the area of the given polygon

**Solution :**

By drawing a horizontal line (FG) parallel to the side DC, we have divided the given polygon into two rectangles.

(i) ABFG is a rectangle

(ii) EGDC is also a rectangle

Area of ABFG :

Area of rectangle = Length x width

length (AF) = 8 ft and width AB = 5 ft

= 8 x 5 = 40 square feet ---(1)

Area of EGDC :

length (DC) = 7 ft and width DE = 3 ft

= 7 x 3 = 21 square feet ---(2)

(1) + (2)

Area of given polygon

= Area of rectangle ABFG + Area of rectangle EGDC

= 40 + 21 = 61 square feet

**Example 2 :**

Find the area of the given polygon

**Solution :**

By drawing a horizontal line (DE) parallel to the side GF, we have divided the given polygon into two rectangles.

(i) ABCE is a rectangle

(ii) DEGF is also a rectangle

Area of rectangle ABCE :

Area of rectangle = Length x width

length (AC) = 20 yd and width AB = 15 yd

= 20 x 15 = 300 square yd ---(1)

Area of DEGF :

length (DG) = 13 yd and width GF = 11 yd

= 13 x 11 = 143 square yd ---(2)

(1) + (2)

Area of given polygon

= Area of rectangle ABCE + Area of rectangle DEGF

= 300 + 143 = 443 square yd

**Example 3 :**

**Find the area of the given polygon**

**Solution : **

By drawing a horizontal line, we have divided the given shape as two parts.

(1) BECF is a rectangle

(2) ABD is triangle

Area of the given polygon

= Area of rectangle BECF + Area of triangle ABD

Area of rectangle BECF :

length CF = 16 cm and width BC = 7 cm

= length x width

= 16 x 7

= 112 cm² ----(1)

Area of triangle ABD :

Base BD = BE - DE => 16 - 8 => 8 cm

Height AB = AC - BC => 13 - 7 => 6

Area of triangle ABD = (1/2) x b x h

= (1/2) x 8 x 6 ==> 24 cm²----(2)

(1) + (2)

Area of the given polygon = 112 + 24 ==> 136 cm²

**Example 4 :**

Find the area of the given polygon

**Solution : **

By drawing a horizontal line, we have divided the given shape as two rectangles.

(1) ABCD is a rectangle

(2) CEFG is rectangle

Area of the given polygon

= Area of rectangle ABCD + Area of triangle DEFG

Area of rectangle ABCD :

length AB = 20 ft and

width AC = AG - CG => 60- 30 = 30 ft

= length x width

= 20 x 30

= 600 ft² ----(1)

Area of triangle DEFG :

length GF = 60 ft and

width FE = 30 ft

= length x width

= 60 x 30

= 1800 ft² ----(1)

(1) + (2)

Area of the given polygon = 600 + 1800 ==> 2400 ft²

**Example 5 :**

Find the area of the given polygon

**Solution :**

Extend the top edge and the right edge of the polygon.

By subtracting the area of triangle BFB from the rectangle ABCD. We can find the area of the given polygon.

Area of the given polygon

= Area of rectangle ABCD - Area of triangle GFB

Area of rectangle ABCD :

length AD = 36 inches and

width AB = 18 inches

= length x width

= 36 x 18

= 648 in² ----(1)

Area of triangle GFB :

base FG = 9 inches and

Height FB = AB - AF ==> 36 - 18 ==> 18 inches

= (1/2) x base x height

= (1/2) x 9 x 18

= 81 in² ----(1)

Area of the given polygon = 648 - 81 = 567 in²

- Area and polygons
- Inverse operations
- Area of square and rectangles
- Area of quadrilaterals
- Area of a parallelogram
- Finding the area of a trapezoid
- Finding the area of a rhombus
- Area of triangles
- Finding the area of a triangle
- Problems using area of a triangles
- Solving area equations
- Writing equations using the area of a trapezoid
- Solving multistep problems
- Area of polygons
- Finding areas of polygons
- Real world problems involving area and perimeter of polygon

After having gone through the stuff given above, we hope that the students would have understood "Area of compound figures".

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