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

Solution 1: 
(i) Length (l) = 46 cm
Breadth (b) = 25 cm
∴ Area of rectangle = (l × b)
= 46 × 25
= 1150 cm²

(ii) Length (l) = 9 m
Breadth (b) = 6 m
∴ Area of rectangle = (l × b)
= 9 × 6
= 54 m²

(iii) Length (l) = 14.5 m
Breadth (b) = 6.8 m
∴ Area of rectangle = (l × b)
= 14.5 × 6.8 m²
= 98.6 m²

(iv) Length (l) = 2 m 5 cm
= 2 × 100 cm + 5 cm
= 200 cm + 5 cm
= 205 cm
Breadth (b) = 60 cm
∴ Area of rectangle = (l × b)
= 205 cm × 60 cm
= 12300 cm²

Solution 2: 
Side of a square plot = 16 m
∴ Area of the square = (side × side) unit²
= (16 × 16) m²
= 256 m²

Solution 3: 
Length of top of table (l) = 2 m 25 cm
= (2 + 0.25) m (100 cm = 1 m)
= 2.25 m
Breadth of top of table (b) = 1 m 20 cm
= (100 + 0.20) m (∵ 100 cm = 1 m)
= 1.20 m
∴ Area of the top of the table = (l × b) sq. unit
= (2.25 × 1.20) m²
= 2.7 m²

Solution 4: 
Length of the sheet of paper = 3 m 24 cm
= 300 cm + 24 cm
= 324 cm
Breadth of the sheet of the paper = 1 m 72 cm
= 100 cm + 72 cm
= 172 cm
∴ Area of the sheet of paper = (324 × 172) cm²
= 55728 cm²
Length of the piece of paper required to make 1 envelope = 18 cm
Breadth of the piece of paper required to make 1 envelope = 12 cm
∴ Area of the piece of paper required to make 1 envelope = (18 × 12) cm²
= 216 cm²
∴ Number of envelopes that can be made = Area of the sheet / Area of paper required to make 1 envelope
= 55728 / 216
= 258 envelopes

Solution 5: 
Length of the room (l) = 12.5 m
Breadth of the room (b) = 8 m
∴ Area of the room = (l × b)
= (12.5 × 8) m²
= 100 m²
Side of square carpet = 8 m
∴ Area of carpet = Side × Side
= 8 m × 8 m
= 64 m²
So, Area left without carpet = Area of the room - Area of the carpet
= (100 - 64)
= 36 m²

Solution 6: 
Length of the lane (L) = 150 m
Breadth of the lane (B) = 9 m
∴ Area of the lane = (L × B) sq. unit
= (150 × 9) m²
= 1350 m²
Now,
Length of one brick (l) = 22.5 cm
Breadth of one brick (b) = 7.5 cm
∴ Area of one brick = (l × b) sq. unit
= 22.5 cm × 7.5 cm
= 168.75 cm²
= 168.75 / 10000 m²
= 0.016875 m²
∴ Number of bricks required = Area of the lane / Area of one brick
= 1350 / 0.016875
= 1350 × 100000 / 16875
= 80000

Solution 7: 
Length of the room (l) = 13 m
Breadth of the room (b) = 9 m
∴ Area of the room = (l × b)
= (13 × 9) m²
= 117 m²
Let, length of required carpet be x m.
Width of carpet = 75 cm = 0.75 m
∴ Area of carpet = (x × 0.75) m²
For carpeting the room :
Area covered by the carpet = Area of the room
∴ x × 0.75 = 117
⇒ x = 117 / 0.75
⇒ x = 117 × 100 / 75
⇒ x = 117 × 4 / 3
⇒ x = 39 × 4
⇒ x = 156
So, the length of the carpet is 156 m.
Cost of 1 m carpet = ₹ 90
∴ Cost 156 m carpet = ₹ (156 × 90)
= ₹ 14040

Solution 8: 
Let the length of the rectangular park = 5x metres and the breadth of the rectangular park = 3x metres
Perimeter of the park = 128 m
∴ 2(length + breadth) = 128
⇒ 2 (5x + 3x) = 128
⇒ 2 × 8x = 128
⇒ x = 128 / (2 × 8)
⇒ x = 8
∴ Length of the park = 5 × 8 = 40 m
Breadth of the park = 3 × 8 = 24 m
Area of the park = (length × breadth)
= (40 × 24) m²
= 960 m²

Solution 9: 
Area of rectangle = 540 cm²
Length of the rectangle (l) = 36 cm
Let, breadth = b cm
∴ (l × b) = 540
⇒ 36 × b = 540
⇒ b = 540 / 36
⇒ b = 15
∴ Width (b) = 15 cm
∴ perimeter of the rectangle = 2(l + b) cm
= 2(36 + 15) cm
= 2 × 51 cm
= 102 cm

Solution 10: 
Area of the rectangle = 600 m²
Breadth (b) = 25 m
Let, length = l m
∴ l × b = Area of the rectangle
⇒ l × 25 = 600
⇒ l = 600 / 25
⇒ l = 24
∴ Length (l) = 24 m
And, perimeter = 2(l + b) unit
= 2 (24 + 25) m
= 2 × 49 m
= 98 m

Solution 11: 
Side of the square plot = 64 m
Perimeter of the square plot = 4 × Side
= 4 × 64 m
= 256 m
According to the question
Perimeter of the rectangular plot = Perimeter of the square plot
Length of the rectangular plot = 70 m
Let, breadth of the plot = b m
Now,
Perimeter = 2 × (Length + Breadth)
⇒ 256 = 2 (70 + b)
⇒ 2 (70 + b) = 256
⇒ 70 + b = 256 / 2
⇒ 70 + b = 128
⇒ b = 128 - 70
⇒ b = 58
∴ Breadth (b) = 58 m
Now, area of the rectangular plot = (length × breadth) unit²
= (70 × 58) m²
= 4060 m²
Area of the square plot = Side × Side unit²
= (64 × 64) m²
= 4096 m².
Square plot has the greater area than that of the rectangular plot by
= (4096 - 4060) m²
= 36 m².

Solution 12: 
Total cost of cultivating the rectangular field = ₹ 71400
Rate of cultivating = ₹ 35 per sq. m
∴ Area of the field = Total cost of cultivating the field / Rate of cultivating
= 71400 / 35 m²
= 2040 m²
Width of the field (b) = 40 m
Let, length of the field = l m
∴ (l × b) = 2040
⇒ l × 40 = 2040
⇒ l = 2040 / 40
⇒ l = 51 m
∴ length (l) = 51 m
Perimeter of the field = 2(l + b)
= 2(51 + 40) m
= 182 m
∴ The cost of fencing 1 m of the field = ₹ 50
Cost of fencing 182 m of the field = ₹ (182 × 50)
= ₹ 9100

Solution 13: 
Measure of a marble tile = 12 cm × 10 cm
Area of wall = 8 m × 6 m = 48 m²
Area of one marble tile = 12 cm × 10 cm
= 120 cm²
= 120 / (100 × 100) m²
= 12 / (10 × 100) m²
= 12 / 1000 m²
∴ Total number of marbles tiles = Area of wall / Area of one marble tile
= 48 m² / (12 / 1000) m²
= 48 × 1000 / 12
= 4000

Solution 14: 
Given, diagonal of square = 5√2 cm
We know that, for square
∴ Side = diagonal / √2
= 5√2 / √2 cm
= 5 cm
∴ Area of the square = side × side
= 5 cm × 5 cm
= 25 cm²
OR
Area of the square = 1/2 × (Diagonal)²
= 1/2 × (5√2)² cm²
= 1/2 × 5 × 5 × √2 × √2 cm²
= 1/2 × 25 × 2 cm²

Solution 15: 
Length of carpet (l) = 30 m
Breadth (b) = 80 cm
= 0.80 m
∴ Area of the carpet = (l × b) sq. unit
= (30 × 0.80) m²
= 24 m²
Cost of one square metre = ₹ 150
Total cost of 24.6 sq. metre = 24 × 150
= ₹ 3600

Solutions 16:
(i)
Area of rectangle ABDC = Length × Breadth
= AB × AC
= AB × AE - CE
= 1 × (10 - 2)
= (1 × 8) m²
= 8 m²
Area of rectangle CEFG = Length × Breadth
= CG × GF
= GD + CD × GF
= 8 + 1 × 2
= (9 × 2) m²
= 18 m²
∴ Area of the complete figure = Area of rectangle ABDC + Area of rectangle CEFG
= (8 + 18) m²
= 26 m²

(ii) 
Area of rectangle AEDC = Length × Breadth
= ED × CD
= (12 × 2) m²
= 24 m²
Area of rectangle FJIH = Length × Breadth
= HI × IJ
= (9 × 1) m²
= 9 m²
Area of rectangle ABGF = Length × Breadth
= AB × AF
= FJ - GJ × {EH - (EA + FH)}
= 9 - 7.5 × {10 - (2 + 1)}
= 1.5 × (10 - 3)
= (1.5 × 7) m²
Area of the complete figure = Area of rectangle AEDC + Area of rectangle FJIH + Area of rectangle ABGF
= (24 + 9 + 10.5) m²
= 43.5 m²

(iii)
Area of the shaded portion = Area of the complete figure - Area of the unshaded figure
= Area of rectangle ABCD - Area of rectangle GBFE
= (CD × AD) - (GB × BF)
= (CD × AD) - {GB × (BC - FC)}
= {(12 × 9) - (7.5 × (12 - 2))}
= {(12 × 9) - (7.5 × 10)}
= (108 - 75) m²
= 33 m²

Solution 17:

(i)

Area of square BCDE = side × side
= 3 cm × 3 cm
= 9 cm²
Area of rectangle ABFK = Length × Breadth
= AK × AB
= (AL + LK) × (AC - BC)
= (2 + 3) × (4 - 3)
= 5 × 1
= 5 cm²
Area of rectangle MLKJ = Length × Breadth
= MJ × ML
= (3 × 2) cm²
= 6 cm²
Area of rectangle FGHI = Length × Breadth
= GH × HI
= (4 × 2) cm²
= 8 cm²
Area of the figure = Area of rectangle BCDE + Area of rectangle ABFK + Area of rectangle MLKJ + Area of square FGHI
= (9 + 5 + 6 + 8) cm²
= 28 cm²

(ii)

Area of rectangle CEFG = Length × Breadth
= EF × CE
= EF × (EA - AC)
= EF × (2 - 1)
= (5 × 1) cm²
= 5 cm²
Area of rectangle ABDC = Length × Breadth
= AB × BD
= (1 × 2) cm²
= 2 cm²
Area of rectangle HIJG = Length × Breadth
= HI × IJ
= (2 × 1) cm²
= 2 cm²
Area of the figure = Area of rectangle CEFG + Area of rectangle HIJG + Area of rectangle ABDC
= (5 + 2 + 2) cm²
= 9 cm²