RS Aggarwal Solutions: Perimeter and Area (Exercise 12A)

Solution 1: The concept of perimeter is used when you are dealing with the boundary or outer edge of a shape or area.
Here are the correct cases where perimeter would be used:
(i) Framing a photo:  We need the perimeter to know how much material is needed for the frame.
(ii) Fencing a field: We need to measure the boundary to know how much fence need to buy.
(vi) Putting lace around a tablecloth: Lace goes around the edge, so we'll need the perimeter.
(vii) Running along a circular track: The distance we run is the perimeter (or circumference, for a circle).
Final Answer: (i), (iii), (vi), (vii)

Solution 2:
(i) Length of the rectangle (l) = 16.8 cm
And Breadth (b) = 6.2 cm
∴ Perimeter of a rectangle = 2 (l + b)
= 2 (16.8 + 6.2) cm
= 2 × 23
= 46 cm
(ii) Length of rectangle (l) = 2 m 25 cm
= (200 + 25) cm
= 2.25 m
And Breadth (b) = 1 m 50 cm
= (100 + 50) cm
= 1.50 m
∴ Perimeter of a rectangle = 2 (l + b)
= 2 (2.25 + 1.50) m
= 2 × 3.75 m
= 7.50 m
(iii) Length of rectangle (l) = 8 m 5 dm
= (80 + 5) dm
= 85 dm
And Breadth (b) = 6 m 8 dm
= (60 + 8) dm
= 68 dm
∴ Perimeter of a rectangle = 2 (l + b)
= 2 (85 + 68) dm
= 2 × 153 dm
= 306 dm
= 30 m 6 dm

Solution 3: 
Length of rectangular field (l) = 58 m
and breadth (b) = 32 m
∴ Perimeter of a rectangle = 2 (l + b)
= 2 (58 + 32) m
= 2 × 90 m
= 180 m
Rate of fencing = ₹ 25 per m
∴ Total cost of fencing the field = ₹ 25 × 180 = ₹ 4500

Solution 4: 
Perimeter of field = 128 m
Ratio in length and breadth = 5 : 3
Let length (l) = 5x
Then breadth (b) = 3x
Now,
2 (length + breadth) = 128
2 (5x + 3x) = 128
⇒ 5x + 3x = 128/2
⇒ 5x + 3x = 64
⇒8x = 64
⇒x = 64/ 8
⇒ x = 8
Length of the field = 5 × 8 = 40 m
Breadth = 3 × 8 = 24 m

Solution 5:
Cost of fencing a rectangular field = ₹ 35 per m
And Total cost of fencing = ₹ 3850
∴ Perimeter of the field =  Total cost/cost of fencing per meter
= 3850/35
= 110 m
Let the length of the field be x m.
Now 
Perimeter of the field = 110
∴ 2 (x + 23) = 110 { breadth = 23m (Given)}
⇒ x + 23 = 55
⇒ x = 55 - 23
⇒ x = 32
Hence, the length of the field = 32 m

Solution 6: 
Total cost of fencing = ₹ 5280
Rate of fencing = ₹ 40 per m
∴ Perimeter of the field =  total cost/rate of fencing
 = 5280/40
= 132 m
Now ratio in length and breadth = 7 : 4
Let length = 7x
It is given that the perimeter of the field is 132 m. 
∴ 2(7x + 4x) = 132
⇒ 7x + 4x = 132/2
⇒ 11x = 66
⇒ x = 66/11
⇒ x = 6
Length = 7 × 6 = 42 m
Breadth = 4 × 6 = 24 m

Solution 7:
(i) Side of square = 6.5 cm
∴ Perimeter = 4 × side
= 4 × 6.5 cm
= 26 cm
(ii) Side of a square = 5.8 m
∴ Perimeter = 4 × side
= 4 × 5.8 m
= 23.2 m
(iii) Side of a square = 3 m 5 dm
= (30 + 5) dm
= 35 dm
∴ Perimeter = 4 × side
= 4 × 35 dm
= 140 dm
= 140/10 m
= 14 m

Solution 8: 
Total cost of fencing a square field = ₹ 5120
Rate of fencing = ₹ 40 per m
∴ Perimeter of square field = Total cost/Rate of fencing per meter
= 5120/40
= 128 m
Now, Perimeter = 4 × side
⇒128 = 4 × side
⇒ side = 128/4
∴ Lenght of each side = 32m

Solution 9: 
Side of a square field = 21 m
Perimeter = 4 × 21 = 84 m
∴ Perimeter of rectangular field = 84 m
Ratio in length and breadth = 4 : 3
Let length (l) = 4x and breadth (b) = 3x
Therefore,
Perimeter = 2 (l + b)
⇒ 84 = 2 (4x + 3x)
⇒ 4x + 3x = 84/2
⇒ 7x = 42
⇒ x = 42/7
⇒ x = 6
∴ Length = 4 × 6 = 24 m
Breadth = 3 × 6 = 18 m

Solution 10:
(i) Sides of the triangle are 7.8 cm, 6.5 cm and 5.9 cm.  
Perimeter of the triangle = (First side + Second side + Third Side) cm  
= (7.8 + 6.5 + 5.9) cm  
= 20.2 cm  
(ii) In an equilateral triangle, all sides are equal.  
Length of each side of the triangle = 9.4 cm  
∴ Perimeter of the equilateral triangle = (3 × Side) cm  
= (3 × 9.4) cm  
= 28.2 cm  
(iii) Length of two equal sides = 8.5 cm  
Length of the third side = 7 cm  
Perimeter of the triangle = {(2 × Equal sides) + Third side} cm  
= {(2 × 8.5) + 7} cm  
= 24 cm

Solution 11:
(i) Each side of a regular pentagon = 8 cm
∴ Perimeter = 5 × Side
= 5 × 8 cm
= 40 cm
(ii) Each side of an octagon = 4.5 cm
∴ Perimeter = 8 × Side
= 8 × 4.5 cm
= 36 cm
(iii) Each side of a regular decagon = 3.6 cm
∴ Perimeter = 10 × Side
= 10 × 3.6 cm
= 36 cm

Solution 12: 
Solution 12: We know that perimeter of a closed figure or a polygon = Sum of its sides
(i) In the figure, sides of a quadrilateral are 45 cm, 35 cm, 27 cm, 35 cm
Perimeter of the figure = Sum of its sides
= (45 + 35 + 27 + 35) cm
= 142 cm
(ii) Sides of a square are 18 cm, 18 cm, 18 cm, 18 cm
Perimeter of the figure = Sum of its sides
= 18 + 18 + 18 + 18
= 72 cm
(iii)Sides of the polygon given are 16 cm, 4 cm, 12 cm, 12 cm, 4 cm, 16 cm and 8 cm
Perimeter of the figure = Sum of its sides
= (16 + 4 + 12 + 12 + 4 + 16 + 8) cm
= 72 cm