PRACTISE SUMS RELATED TO PERIMETER
CLASS 6TH , 7TH ,8TH
Q.1) Find the perimeter of :-
i) a regular pentagon of side 8cm.
ii) a regular octagon of side 4.5cm .
iii) a regular decagon of side 3.6cm .
SOLUTION:- i) Regular pentagon
Side = 8cm.
No. of sides = 5
Perimeter of pentagon = ?
Perimeter of pentagon = no. of sides * side
P = 5 * 8
P = 40 cm.
Ans :- The perimeter of the pentagon is 40 cm.
——————————————————————————————–
SOLUTION:- ii) Regular Octagon
Side = 4.5 cm.
No. of sides = 8
Perimeter of octagon = ?
Perimeter of octagon = no. of sides * side
P = 8 * 4.5
P = 36 cm.
Ans :- The perimeter of regular octagon is 36cm.
——————————————————————————————–
SOLUTION :- iii) Regular decagon
Side = 3.6 cm.
No. of sides = 10
Perimeter of decagon = ?
Perimeter of decagon = no. of sides * side
P = 10 * 3.6
P = 36 cm.
Ans :- The perimeter of regular decagon is 36 cm.
——————————————————————————————-
Q2). Find the perimeter of :-
i) a triangle of sides 7.8cm ,6.5cm and 5.9cm.
ii) an equilateral triangle of side 9.4cm.
iii) an isosceles triangle with equal sides 8.5cm each and third side 7cm.
SOLUTION :- i) Triangle
Given:
Seg. AB = 7.8cm
Seg. AC = 5.9cm
Seg. BC = 6.5cm
Perimeter of the triangle = P = ?
P = sum of all the sides = AB + AC+ BC
= 7.8 + 5.9 + 6.5 = 20.2cm
SOLUTION:- ii) Equilateral Triangle
Given:
each side = 9.4cm
Perimeter of equilateral triangle = P = ?
P = 3*S
P = 3*9.4
P = 28.2cm
SOLUTION: iii) Isosceles
Given:
Line Segment = Seg.
each equal side
Seg. = AB = AC = 8.5cm
Seg. BC = 7cm
Perimeter of isosceles triangle = P = ?
P = sum of all sides
P = AB + AC + BC
P = 8.5 + 8.5 + 7
P = 24cm
Q.3) Find the cost of fencing a rectangular field 62m long and 33m wide at Rs.16 per meter.
SOLUTION:- Rectangular Field
Given:
Length = L = 62m
Width = Breadth = B = 33m
Rate of fencing = Rs. 16 per meter
Find the cost of fencing = ?
I) Perimeter of rectangular field = L*B
P = 62*33
P = 2046m
II) Cost of fencing = P*Rate
= 2046*16
=32,736
ANS: Rs.32,736 is the cost of fencing.
———————————————————————————————————
Q.4) The length and the breadth of a rectangular field are in the ratio 5:3. If its perimeter is 128m. Find the dimensions of the field.
SOLUTION: Rectangular Field
Given:
Ratio of the length and breadth = 5:3
Assume Length = L = 5x m
Breadth = B = 3x m
Perimeter= 128m
Find the dimensions = ?
i.e., L = ? , B = ?
Perimeter of rectangle = 2*( L+ B )
128 = 2*( 5x + 3x )
128 = 2*(8x)
128/2 = 8x
64 = 8x
64/8 = x
8m = x
Hence, L = 5x = 5*8 = 40m
B = 3x = 3*8 = 24m
——————————————————————————————————
Q.5)The cost of fencing a rectangular field at Rs.18 per meter is Rs.1980. If the width of the field is 23m. Find its length.
SOLUTION: Rectangular Field
Given:
Cost of fencing the field = Rs.1980
Rate of fencing the field = Rs.18 per meter
Width = Breadth = B = 23m
Length = L = ?
I) Per meter Cost
1m Rs.18
? Rs 1980
Hence, total meter of the field fencing = 1980/18 = 110m
Perimeter of the rectangular field = 110m
II) Perimeter of rectangle = 2*( L+ B )
110 = 2*( L + 23 )
110/2 = L + 23
55 = L + 23
55 – 23 = L
32 = L
Length = 32m
ANS: The length of the rectangular field is 32m.
————————————————————————————————————–