Notes on Perimeter | Grade 7 > Compulsory Maths > Perimeter, Area and Volume | KULLABS.COM

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#### Perimeter of Plane Figure

All the Triangle , Circle , Square, and Rectangle are plan figure . The total length of the boundary lines of plan figure is called its perimeter.Perimeter of Triangles :

#### Perimeter of Triangles :

The sum of the length of its three sides is known as a perimeter of triangles . Therefore,

$$\therefore$$ Perimeter of $$\triangle DEF$$ =DE + EF+ FD = a+b+c

The half of its perimeter is known as semi -perimeter of triangle . It is denoted by the letter 's'.

$$\therefore$$ Semi -perimeter of $$\triangle$$DEF= $$\frac{a+b+c}{2}$$

#### Perimeter of Rectangles :

The opposite sides of rectangle are always equal . So , the lengths = EF=GH=l

The perimeter of the rectangles EFGH =EF+FG+GH+HG

=l+b+l+b=2l+2b

=2(l+b)

$$\therefore$$Perimeter of rectangle = 2(l+b)

In the case of Square , its perimeter =2(l+l)

=2x2l=4l

#### Perimeter of circles :

At first draw three circles with radii 2cm , 3cm, and 4cm, and place the pieces of strings along the circumference of each separately .

Then start measuring the length of each string separately with the help of scale . Then, find the ratios of the length of the circumference of each circle to its diameter . Now , you all will find the ratio $$\frac{circumference}{diameter}$$ is almost the same for every circle. The constant ratio is represented by Greek letter 'π' (pie). So , the circumference of circle iscand its diameter bed,

Now , $$\frac{circumfernce }{diameter})=π or, \(\frac{c}{d})=π or, c=πd Diameter of circle (d) =2xradius (r). So, circumference or the perimeter of circle (c) =2πr The perimeter of circle =πd or2πr • All the Triangle , Circle , Square, and Rectangle are plan figure . The total length of the boundary lines of plan figure is called its perimeter. • The sum of the length of its three sides is known as a perimeter of triangles . • The opposite sides of a rectangle are always equal . . ### Questions and Answers #### Click on the questions below to reveal the answers Solution: Here, the length of the field (l) = 25 m the perimeter of the field = 86 m Now, the perimeter of the rectangular field = 86 m or, 2(l + b) = 86 m or, 25 + b = \(\frac{86}{2}$$ m
or, 25 + b = 43 m
or, b = (43 - 25) m
or, b = 18 m
∴ the required breadth of the field is 18 m

Solution:

Here,
the radius of the circular ground (r) = 35m
its perimeter = 2πr
= 2 × $$\frac{22}{7}$$ × 35 m
= 2 × 22 × 5
= 220 m
∴ The required length of wire = 4 × 220m
= 880 m
Again,
the required cost of fencing = 880 × Rs 10
= Rs 8800

0%

l + b
2 l + b
4 l
2(l + b)

(πr)2
πr
πr2
2πr
• ### What is the formula of a semi-perimeter of a triangle?

a + b + c
(frac{a + b - c}{2})
(frac{a + b + c}{2})
(frac{a + b + c}{3})

18m
13m
21m
15m

73m
56m
43m
24m

42m
19m
38m
12m

12.4 cm
18.3 CM
16.2 cm
11.2 cm

25.4 m
26.5 m
21.3 m
23.6 m

s
t
s-p
p
• ### The value of "π" is ______ .

(frac{7}{22})
22
(frac{2.2}{7})
(frac{22}{7})

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