Let’s learn how to find the Perimeter and Area of Mixed Figures. We will get mixed figures with the combination of several shapes. The mixed figures may be a combination of any shapes like triangle and square, square and rectangle, triangle and pyramid, etc. This article completely explains the formulas for perimeter and area of the different mixed figures, steps on how to find the area and perimeter of combined figures. 9th Grade Math students can get different examples with their explanations step by step with relevant images for better understanding.

## Mixed Figures â€“ Definition

When we add different shapes to form a combined shape, we can see the different shapes. The shape which is the combination of different shapes is known as mixed figures. Below is an example of a mixed figure.

From the above figure, ABCD is a rectangle and ABE is a triangle. By combining both figures, we got a new figure ABCDE.

### How to find Perimeter and Area of Mixed Figures?

Follow the below procedure to find the perimeter and area of mixed figures. The step-by-step process to find the area and perimeter is given separately below.

#### How to Find Area of Mixed Figures?

Step 1: Firstly, find out the shape.

Step 2: If you identify the individual shapes then find the area separately.

Step 3: Then, add the total area to find the complete answer.

Step 4: Finally, note down the answer in square units.

Also, check:

- Perimeter and Area of Irregular Figures
- Perimeter and Area of Square
- Perimeter and Area of Rectangle

#### How to Find the Perimeter of Mixed Figures?

The perimeter of the mixed figures can find by adding the total side lengths of the shape. If rectangle and square are two shapes then add the lengths of four sides.

### Perimeter and Area of Mixed Shapes Examples

**Example 1:** The length and breadth of a rectangular field are 16 cm and 12 cm respectively. Two right-angled isosceles triangles are constructed outside the rectangular field, with the longer sides as the hypotenuses. On the 12 cms sides of the rectangular field, two equilateral triangles are constructed outside. Find the total area and perimeter of the figure.

**Solution:** The figure consists of the following.

(i) The rectangular field ABCD, whose area = 16 Ã— 12 cm^{2 }Â = 192 cm^{2 }

(ii) Two equilateral triangles ADH and BCG. For each, area = âˆš3/4 Ã— 12 cm^{2 }Â = 3 âˆš3 cm^{2}

(iii) Two isosceles right-angled triangles CDE and ABF, whose areas are equal.

According to Pythagorasâ€™ theorem, if CE = ED = x then x^{2}Â + x^{2}Â = (16)^{2}Â cm^{2}Â or 2x^{2}Â = 256 cm^{2}. Therefore, x = 8âˆš2 cm.

Therefore, the area of the âˆ†CDE = 1/2 CE Ã— DE = 1/2 x^{2}Â = 1/2 (8âˆš2)^{2} cm^{2} = 64 cm^{2}

Therefore, area of the figure = area of the rectangular field ABCD + 2 Ã— area of the âˆ†BCG + 2 Ã— area of the âˆ†CDE

= (192 + 2 Ã— 3 âˆš3 + 2 Ã— 64) cm^{2} = 1920âˆš3 cm^{2} =

The perimeter of the figure = length of the boundary of the figure

= AF + FB + BG + GC + CE + ED + DH + HA

= 4 Ã— CE + 4 Ã— BG

= (4 Ã— 8âˆš2 + 4 Ã— 12) cm

= 93.12 cm

**Example 2.** The dimension of a field is 220 m Ã— 160 m. The field is to be converted into a garden after some days, leaving a path 10 m broad around the garden. Find the total cost of making the new garden if the cost per square meter is Rs 24.

**Solution:
**

For the garden, length = (220 â€“ 2 Ã— 10) m = 200 m, and Breadth = (160 â€“ 2 Ã— 10) m = 140 m.

Therefore, area of the garden = 200 m Ã— 140 m = 28000 m

^{2}.

Therefore, the total cost of making the garden = 28000 Ã— Rs 24 = 672000

**Example 3.** A square-shaped piece of paper is made into two pieces along a line joining a corner and a point on an opposite edge. If the ratio of the areas of the two pieces is 6:2, find the ratio of the perimeters of the smaller piece and the original piece of paper.

**Solution:**

Let ABCD be the square-shaped piece of paper. Let its side measure a unit.

It is cut along AE. Let DE = q units

Area of the âˆ†EDA = 1/2 (AD Ã— DC) = 1/2 pqÂ square units.

Area of the square ABCD = p^{2} square units.

According to the question,

(Area of the quadrilateral ABCE/ Area of the Î”EDA) = 6/2

(Area of the quadrilateral ABCE/ Area of the Î”EDA) + 1 = 8/2

(area of the quadrilateral ABCE + area of the Î”EDA)/area of the Î”EDA = 8/2

(Area of the quadrilateral ABCD/ Area of the Î”EDA) = 4

p^{2}/(pq/2) = 4

2p/q = 4

p = 2q

q = p/2

According to Pythagorasâ€™ theorem,

AE^{2} = AD^{2} + DE^{2}

Therefore, AE^{2} = p^{2} + q^{2}

= p^{2} + (p/2)^{2}

= p^{2} + 1/4 (p^{2})

= 5/4 (p^{2})

Therefore, AE^{2} = âˆš5/2p.

Now, the perimeter of the Î”EDA/perimeter of the square ABCD = (ED + AD + AE)/4p

= (p/2 + p + âˆš5p/2)/4p

= [(3+âˆš5)p/2]/4p

= (3+âˆš5)/8

= (3+âˆš5) : 8.

**Example 4.** From a 40 cm Ã— 20 cm board having an F-shaped block is cut out as given in the below figure. What is the area of the face of the remaining board? Also, find the length of the boundary of the board or block.

**Solution:**

The given figure is the combination of three rectangular blocks.

Therefore, area of a face of the block = 40 Ã— 6 cm^{2} + 6 Ã— 4 cm^{2} + 14 Ã— 6 cm^{2}

= 240 cm^{2} + 28 cm^{2} + 84 cm^{2}

= 352 cm^{2}

Area of a face of the uncut board = 40 Ã— 20 cm^{2}

= 800 cm^{2}

Therefore, the area of the face of the remaining board = 800 cm^{2} – 352 cm^{2}Â = 448 cm^{2}

Required length of the boundary = (40 + 6 + 22 + 4 + 6 + 4 + 6 + 14 + 6 + 20) cm = 128 cm