A Combined shape is the combination of several closed shapes. The perimeter is calculated by adding all the outer sides (boundaries) of the combined shape.

**Perimeter
and Area of Combined Shapes**

A ** Combined shape**
is the combination of several closed shapes. The perimeter is calculated by adding
all the outer sides (boundaries) of the combined shape. The area is calculated by
adding all the areas of closed shapes from which the combined shape is formed.

** **

__Example 12 __

Find the perimeter of the given figure.

**Solution **

Perimeter = Total length of the boundary

= (6 + 2 + 10 + 3 + 2 + 1 + 3 + 4
+ 2 + 6 + 9) *cm*

= 48 *cm*

** **

__Example 13____ __

Find the perimeter and the area of the following
‘L’ shaped figure.

**Solution**

Perimeter
= (28 + 7 + 21 + 21 + 7 + 28)* cm.*

= 112* cm.*

To find the area of the L shaped figure, it is divided
into two rectangles A and B.

The area of the ‘L’ shaped figure = (196 + 147)
*sq*. *cm*

= 343 *sq*. *cm*.

** **

**Activity **

Find the area of the given ‘L’ shaped
rectangular figure by dividing it into squares of equal sizes.

**Think**

Can you find the area of ‘L’ shaped figure
as the difference between two areas.

**Try these**

**Measure using ruler and find the perimeter
of each of the following diagram.**

**Activity**

Form all possible shapes of perimeter
80 *cm* with 9 identical squares, each of
side 4 *cm*.

__1.
Impact on Removing / Adding a portion from / to a given shape__

Consider a rectangle of sides 8 *cm* and 12 *cm*.

Length, *l* = 12 *cm*; Breadth *b* = 8* cm.*

Area, A = (*l* × *b*) sq. units.

= 12 × 8

= 96 sq.* cm.*

Perimeter, P = 2 (*l* + *b*) units.

= 2 (12 + 8)

= 40 *cm*.

Find the area and perimeter of the rectangle in
the following situations and observe the changes.

**Situation 1**

A square of side 3* cm *is cut at a corner of the rectangle.

Area, A = (*l* × *b*) – (s × s) *sq*. units.

= (12 × 8) – (3 × 3)

= 87 *sq*. *cm*.

Perimeter, P = (Total boundary) units.

= 8+12+5+3+3+9 = 40 *cm*.

The perimeter is not changed. But the area is reduced.

**Situation 2**

A square of side 3 *cm* is attached to the rectangle.

Area, A = (*l* × *b*) + (s × s) *sq*. units.

= (12 × 8) + (3 × 3)

= 105 *sq*. *cm*.

Perimeter, P = (Total boundary) units.

= 8+12+11+3+3+9 = 46 *cm*.

Here both the perimeter and the area are increased.

** **

__Example 14 __

Four identical square floor mats of
side 15** ***cm*** **are joined together to form either a
rectangular mat or a square mat. Which mat will have a larger area and a longer
perimeter?

**Solution**

Perimeter of a rectangle, P = 2 (*l* + *b*) units.

= 2 (60+15) *cm*. = 150 *cm*.

Area of a rectangle, A = (*l* × *b*) sq. units.

= 60 × 15 = 900 *sq*. *cm*.

Perimeter of a square, P = (4 × *s*) units

= (4 × 30) *cm* = 120 *cm*

Area of a square, A = (*s* × *s*) sq.
units.

= 30 × 30 = 900 *sq*. *cm*.

There is no change in their areas. But, the rectangular mat will have longer perimeter.

** **

**Activity**

Cut a rectangular sheet along one
of its diagonals. Two identical scalene right angled triangles are obtained. Join
them along their** **sides of identical length in all possible ways. Six different
shapes** **can be obtained. Four of them are given. Find the remaining two**
**shapes. Find the perimeter of all the six shapes and fill in the table.

Based on the above activity answer
the following questions:

i) Are the perimeters same for all
the shapes?

ii) Which shape has the longest perimeter?

iii) Which shape has the shortest
perimeter?

iv) Are the areas of all the shapes
same? Why?

**DO YOU KNOW?**

• Shapes with the same perimeter may
have different areas.

• Shapes with the same area may have
different perimeters.

Tags : Term 3 Chapter 3 | 6th Maths , 6th Maths : Term 3 Unit 3 : Perimeter and Area

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6th Maths : Term 3 Unit 3 : Perimeter and Area : Perimeter and Area of Combined Shapes | Term 3 Chapter 3 | 6th Maths

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