New Learning Composite Mathematics Class 5 S.K. Gupta Anubhuti Gangal Perimeter and Area Chapter 17A Solution
Self-Practice 17A perimeter.
1) Find the perimeter of a square with side length
a) 15 cm b) 8 mm c) 1.9 mm d) 2.13 m e) 8.2 cm f) 0.14 km
Ans:
a) The perimeter of a square of side 15 cm is = 4×15 cm = 60 cm
b) The perimeter of a square of side 8 mm is = 4×8 mm = 32 mm
c) The perimeter of a square of side 9 mm = 4×1.9 mm = 7.6 mm
d) The perimeter of a square of side 2.13 m = 4×2.13 m = 8.52 m
e) The perimeter of a square of side 8.2 cm = 4×8.2 cm = 32.8 cm
2) Find the perimeter of a rectangle whose dimensions are given below.
(a) (b) (c) (d) (e) (f)
Length | 10 cm | 18 m | 7 mm | 6.4 cm | 12.5 m | 12.5 km |
Breadth | 5 cm | 12 m | 2.5 mm | 36 mm | 70 cm | 3.2 km |
Ans:
a) The perimeter of the rectangle = 2×(10+5) cm. = 30 cm.
b) The perimeter of the rectangle = 2×(18+12) m = 60 m
c) The perimeter of the rectangle = 2×(7+2.5) mm = 19 mm
d) The perimeter of the rectangle = 2×(6.4+3.6) cm = 20 cm
e) The perimeter of the rectangle = 2×(12.5+0.7) m = 26.4 m
f) The perimeter of the rectangle = 2× (12.5+3.2) km = 31.4 km
3) How many metre of fencing is required to enclose a rectangular vegetable garden 6.8 m long and 5.6 m wide?
Ans: The perimeter of the garden is 2×(6.8+5.6) m = 24.8 m
So the length of the fencing required to enclose the garden is 24.8 m.
4) The length of a rectangular park is 0.140 km and its breadth is 80 m. What is the perimeter of the park in meters? Three rounds of barbed wire are needed to fence the park. Find the cost of fencing the park if the barbed wire costs ₹2.20 per metre.
Ans: length = 0.140 km = 140 m, breadth = 80 m.
The perimeter of the park is 2×(140+80) = 440 m.
Three round of barbed wire is requited so the total length required to cover the park is 3×440 m = 1320 m
The cost of total wire = ₹ (1320×2.2) = ₹1904.
5) How much lace will be needed to put around a square cushion of side 70 cm? What will be the cost of the lace required if the lace costs ₹10 per metre?
Ans: The perimeter of the cushion = 4×70 cm = 280 cm = 0.28 m
Cost of the lace required = ₹10×0.28 = ₹2.8
6) Complete the given table.
(a) (b) (c) (d)
Length |
20 mm | |||
Breadth |
14 m | 8 cm |
12 cm |
|
Perimeter | 58 m | 34 cm | 50 mm |
40 cm |
Ans:
a) length + breadth = Perimeter/2
or, length + 14 = 58/2
or, length = 29-14 = 15
length is 15 m
b) length + breadth = Perimeter/2
or, length + 8 = 34/2
or, length = 17 – 8 = 9
length = 9 cm
c) length + breadth = Perimeter/2
or, 20+ breadth = 50/2
or, breadth = 25 – 20 = 5
breadth = 5 mm
d) length + breadth = Perimeter/2
or, length +12 = 40/2
or, length = 20 – 12 = 8
length = 8 cm
7) Find the length of a side of a square whose perimeter is
a) 20.8 cm b) 12.6 cm c) 24 m d) 7.2 m
Ans:
a) The length of the side of the square is = 20.8/4 cm = 5.2 cm
b) The length of the side of the square is = 12.6/4 cm = 3.15 cm
c) The length of the side of the square is = 24/4 m = 6 m
d) The length of the side of the square is = 7.2/4 m = 1.8 m
8) A rope of length 40.8 m was used to fence off a square patch of land for a garden. Find the length of each side of the square garden.
Ans: According to the question 40.8 m rope was used to fence off a square land.
So the perimeter of the land is 40.8 m.
So the length of the side of the square land is = 40.8/4 m = 10.2 m
9) The perimeter of an envelope is 38 cm. If the breadth of the envelope is 5 cm, what is its length?
Ans: the envelop is a rectangle
So, we know that for a rectangle
length + breadth = Perimeter/2
or, length + 5 = 38/2
or, length = 19 – 5 = 14
length of the envelop is 14 cm
10) The figures given on the right show a rectangle and a square with dimension given.
(a) What is the perimeter of each figure?
(b) Which figure has greater perimeter and by how much?
Ans:
a) the first figure is a rectangle with length 110 cm and breadth 60 cm.
The perimeter of the rectangle is 2×(60+110) cm = 340 cm.
The second figure is a square with sides 70 cm.
The perimeter of the square is 4×70 cm = 280 cm.
b) The rectangle has greater perimeter.
The difference between their perimeter is (340 – 280) cm = 60 cm
See also :
New learning Composite Mathematics class 5 S.K Gupta & Anubhuti Gangal complete solution