Rs Aggrawal 2019 2020 Solutions for Class 6 Math Chapter 21 Concept Of Perimeter And Area are provided here with simple step-by-step explanations. These solutions for Concept Of Perimeter And Area are extremely popular among Class 6 students for Math Concept Of Perimeter And Area Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggrawal 2019 2020 Book of Class 6 Math Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggrawal 2019 2020 Solutions. All Rs Aggrawal 2019 2020 Solutions for class Class 6 Math are prepared by experts and are 100% accurate.
Page No 222:
Answer:
We know: Perimeter of a rectangle =
(i) Length = 16.8 cm
Breadth = 6.2 cm
Perimeter =
=
(ii) Length = 2 m 25 cm
=(200+25) cm (1 m = 100 cm )
= 225 cm
Breadth =1 m 50 cm
= (100+50) cm (1 m = 100 cm )
= 150 cm
Perimeter =
=
(iii) Length = 8 m 5 dm
= (80+5) dm (1 m = 10 dm )
= 85 dm
Breadth = 6 m 8 dm
= (60+8) dm (1 m = 10 dm )
= 68 dm
Perimeter =
=
Page No 222:
Answer:
Length of the field = 62 m
Breadth of the field = 33 m
Perimeter of the field = 2(l + b) units
= 2(62 + 33) m =190 m
Cost of fencing per metre = Rs 16
Total cost of fencing = Rs (16190) = Rs 3040
Page No 222:
Answer:
Let the length of the rectangle be 5x m.
Breadth of the rectangle = 3x m
Perimeter of the rectangle = 2(l + b)
= 2(5x + 3x) m
= (16x) m
It is given that the perimeter of the field is 128 m.
Page No 222:
Answer:
Total cost of fencing = Rs 1980
Rate of fencing = Rs 18 per metre
Perimeter of the field =
Let the length of the field be x metre.
Perimeter of the field = 2(x + 23) m
Hence, the length of the field is 32 m.
Page No 222:
Answer:
Total cost of fencing = Rs 3300
Rate of fencing = Rs 25/m
Perimeter of the field =
Let the length and the breadth of the rectangular field be 7x and 4x, respectively.
Perimeter of the field = 2(7x + 4x) = 22x
It is given that the perimeter of the field is 132 m.
Page No 222:
Answer:
(i) Side of the square = 3.8 cm
Perimeter of the square = (4side)
= (43.8) = 15.2 cm
(ii) Side of the square = 4.6 cm
Perimeter of the square = (4side)
= (44.6) = 18.4 cm
(iii) Side of the square = 2 m 5 dm
= (20+5) dm (1 m = 10 dm)
= 25 dm
Perimeter of the square = (4side)
= (425) = 100 dm
Page No 222:
Answer:
Total cost of fencing = Rs 4480
Rate of fencing = Rs 35/m
Perimeter of the field =
Let the length of each side of the field be x metres.
Perimeter = (4x) metres
Hence, the length of each side of the field is 32 m.
Page No 222:
Answer:
Side of the square field = 21m
Perimeter of the square field = (421) m
= 84 m
Let the length and the breadth of the rectangular field be 4x and 3x, respectively.
Perimeter of the rectangular field = 2(4x + 3x) = 14x
Perimeter of the rectangular field = Perimeter of the square field
Page No 222:
Answer:
(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 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
Page No 222:
Answer:
(i) Length of each side of the given pentagon = 8 cm
Perimeter of the pentagon = (58) cm
= 40 cm
(ii) Length of each side of the given octagon = 4.5 cm
Perimeter of the octagon = (84.5) cm
= 36 cm
(iii) Length of each side of the given decagon = 3.6 cm
Perimeter of the decagon = (103.6) cm
= 36 cm
Page No 222:
Answer:
(i) Perimeter of the figure = Sum of all the sides
=(27 + 35 + 35 + 45) cm
= 142 cm
(ii) Perimeter of the figure = Sum of all the sides
=(18 + 18 + 18 + 18) cm
= 72 cm
(iii) Perimeter of the figure = Sum of all the sides
=(8 + 16 + 4 + 12 + 12 + 16 + 4) cm
= 72 cm
Page No 224:
Answer:
(i) Radius, r = 28 cm
(ii) Radius, r = 10.5 cm
(iii) Radius, r = 3.5 m
Page No 224:
Answer:
(i)
(ii)
(iii)
Page No 224:
Answer:
Let the radius of the given circle be r cm.
Circumference of the circle = 176 cm
Circumference =
Page No 224:
Answer:
Page No 224:
Answer:
Radius of the wheel =
Circumference of the wheel =
In 1 revolution the wheel covers a distance equal to its circumference.
Page No 224:
Answer:
Page No 226:
Answer:
The figure contains 12 complete squares.
Area of 1 small square = 1 sq cm
∴ Area of the figure = Number of complete squares Area of the square
=
=12 sq cm
Page No 226:
Answer:
The figure contains 18 complete squares.
Area of 1 small square = 1 sq cm
∴ Area of the figure = Number of complete squares Area of the square
=
=18 sq cm
Page No 226:
Answer:
The figure contains 14 complete squares and 1 half square.
Area of 1 small square = 1 sq cm
∴ Area of the figure = Number of squares Area of the square
=
= sq cm
Page No 226:
Answer:
The figure contains 6 complete squares and 4 half squares.
Area of 1 small square = 1 sq cm
∴ Area of the figure = Number of squares Area of the square
=
=8 sq cm
Page No 226:
Answer:
The figure contains 9 complete squares and 6 half squares.
Area of 1 small square = 1 sq cm
∴ Area of the figure = Number of squares Area of the square
=
=12 sq cm
Page No 226:
Answer:
The figure contains 16 complete squares.
Area of 1 small square = 1 sq cm
∴ Area of the figure = Number of squares Area of a square
=
=16 sq cm
Page No 226:
Answer:
In the given figure, there are 4 complete squares, 8 more than half parts of squares and 4 less than half parts of squares.
We neglect the less than half parts and consider each more than half part of the square as a complete square.
∴ Area = (4 + 8) sq cm
= 12 sq cm
Page No 226:
Answer:
In the given figure, there are 9 complete squares, 5 more than half parts of squares and 7 less than half parts of squares.
We neglect the less than half parts of squares and consider the more than half squares as complete squares.
∴ Area of the figure = (9 + 5) sq cm
= 14 sq cm
Page No 226:
Answer:
The figure contains 14 complete squares and 4 half squares.
Area of 1 small square = 1 sq cm
Area of the figure = Number of squares Area of one square
=
=16 sq cm
Page No 229:
Answer:
(i) Length = 46 cm
Breadth = 25 cm
Area of the rectangle = (Length Breadth) sq units
= (4625) cm2 = 1150 cm2
(ii) Length = 9 m
Breadth = 6 m
Area of the rectangle = (Length Breadth) sq units
= (96) m2 = 54 m2
(iii) Length = 14.5 m
Breadth = 6.8 m
Area of the rectangle = (Length Breadth) sq units
= () m2 = m2 =98.60 m2
(iv) Length = 2 m 5 cm
= (200+5) cm (1 m = 100 cm )
=205cm
Breadth = 60 cm
Area of the rectangle = (Length Breadth) sq units
= (20560) cm2 = 12300 cm2
(v) Length = 3.5 km
Breadth = 2 km
Area of the rectangle = (Length Breadth) sq units
= (3.52) km2 = km2 =7 km2
Page No 230:
Answer:
Side of the square plot = 14 m
Area of the square plot = (Side)2 sq units
= (14)2 m2
= 196 m2
Page No 230:
Answer:
Length of the table = 2 m 25 cm
= (2 + 0.25) m (100 cm = 1 m)
= 2.25 m
Breadth of the table = 1 m 20 cm
= (1 + 0.20) m (100 cm = 1 m)
=1.20 m
Area of the table = (Length × Breadth) sq units
= (2.25 × 1.20) m2
= m2
= 2.7 m2
Page No 230:
Answer:
Length of the carpet = 30 m 75 cm
=(30 + 0.75) cm (100 cm = 1 m)
= 30.75 m
Breadth of the carpet = 80 cm
= 0.80 m (100 cm = 1 m)
Area of carpet = ( Length breadth ) sq units
Cost of 1 m2 carpet= Rs 150
Cost of 24.6 m2 carpet = Rs (24.6150)
= Rs 3690
Page No 230:
Answer:
Length of the sheet of paper = 3 m 24 cm = 324 cm
Breadth of the sheet of paper = 1 m 72 cm = 172 cm
Area of the sheet = (Length Breadth)
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 = (1812) cm2
= 216 cm2
Page No 230:
Answer:
Length of the room = 12.5 m
Breadth of the room = 8 m
Area of the room = (LengthBreadth)
Side of the square carpet = 8 m
Area of the carpet = (Side)2
= 82 m2
= 64 m2
Area of the floor which is not carpeted = Area of the room − Area of the carpet
= (100 − 64) m2
= 36 m2
Page No 230:
Answer:
Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (LengthBreadth)
Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (LengthBreadth)
Page No 230:
Answer:
Length of the room = 13 m
Breadth of the room = 9 m
Area of the room = (139) m2 = 117 m2
Let length of required carpet be x m.
Breadth of the carpet = 75 cm
= 0.75 m (100 cm = 1 m)
Area of the carpet = (0.75x) m2
= 0.75x m2
For carpeting the room:
Area covered by the carpet = Area of the room
So, the length of the carpet is 156 m.
Cost of 1 m carpet = Rs 65
Cost 156 m carpet = Rs (15665)
= Rs 10140
Page No 230:
Answer:
Let the length of the rectangular park be 5x.
∴ Breadth of the rectangular park = 3x
Perimeter of the rectangular field = 2(Length + Breadth)
=2(5x + 3x)
= 16x
It is given that the perimeter of rectangular park is 128 m.
Area of the park = (Length Breadth) sq units
Page No 230:
Answer:
Side of the square plot = 64 m
Perimeter of the square plot =
Area of the square plot = (Side)2
= 642 m2
= 4096 m2
Let the breadth of the rectangular plot be x m.
Perimeter of the rectangular plot = 2(l+b) m
= 2(70+x) m
Perimeter of the rectangular plot = Perimeter of the square plot (Given)
So, the breadth of the rectangular plot is 58 m.
Area of the rectangular plot =
Area of the square plot − Area of the rectangular plot
= (4096 − 4060)
= 36 m2
Area of the square plot is 36 m2 greater than the rectangular plot.
Page No 230:
Answer:
Total cost of cultivating the field = Rs 71400
Rate of cultivating the field = Rs 35/m2
Let the length of the field be x m.
Perimeter of the field = 2(l+b)
= 2(51+40) m
= 182 m
Cost of fencing 1 m of the field = Rs 50
Cost of fencing 182 m of the field = Rs (18250)
= Rs 9100
Page No 230:
Answer:
Let the width of the rectangle be x cm.
Length of the rectangle = 36 cm
Area of the rectangle = () = () cm2
It is given that the area of the rectangle is 540 cm2.
Perimeter of the rectangle = 2(Length + Width) cm
= 2(36 + 15) cm
= 102 cm
Page No 230:
Answer:
Length of the wall = 4 m = 400 cm
Breadth of the wall = 3 m = 300 cm
Area of the wall = (400×300) cm2 = 120000 cm2
Length of the tile = 12 cm
Breadth of the tile = 10 cm
Area of one tile = (12×10) cm2 = (120) cm2
Cost of 1 tile = Rs 22.50
Cost of 1000 tiles = (1000 × 22.50) = Rs 22500
Thus, the total cost of the tiles is Rs 22500.
Page No 230:
Answer:
Let the length of the rectangle be x cm.
Breadth of the rectangle is 25 cm.
Area of the rectangle = (Length × Breadth) cm2
= (x×25) cm2
=25x cm2
It is given that the area of the rectangle is 600 cm2.
So, the length of the rectangle is 24 cm.
Perimeter of the rectangle = 2(Length + Breadth) units
= 2(25 + 24) cm
= 98 cm
Page No 230:
Answer:
Area of the square =
Page No 230:
Answer:
(i) Area of rectangle ABDC = Length Breadth
= ABAC (AC = AE − CE)
=
= 8 m2
Area of rectangle CEFG = Length Breadth
= CGGF (CG = GD + CD)
=
= 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
=
= 24 cm2
Area of rectangle FJIH = Length Breadth
= HI IJ
=
= 9 m2
Area of rectangle ABGF = Length Breadth
= AB AF {(AB = FJ − GJ) and AF = EH − (EA + FH)}
=
= 10.5 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
=(CDAD) − (GBBF)
= (BF = BC − FC)
=(108 − 75) m2
=33 m2
Page No 231:
Answer:
(i) Area of square BCDE= (Side)2
= (CD)2
= (3)2 cm2
= 9 cm2
Area of rectangle ABFK =
= AKAB [(AB = AC − BC) and (AK = AL + LK)
= (51) cm2
= 5 cm2
Area of rectangle MLKG =
= ML MG
= (2 3) cm2
= 6 cm2
Area of rectangle JHGF=
= JHHG
= (24) cm2
= 8 cm2
Area of the figure = Area of rectangle ABFK + Area of rectangle MLKG + Area of rectangle JHGF + Area of square BCDE
= (9 + 5 + 6 + 8) cm2
= 28 cm2
(ii) Area of rectangle CEFG=
= EFCE
= (15) cm2 (CE = EA − AC)
= 5 cm2
Area of rectangle ABDC =
= ABBD
= (12) cm2
= 2 cm2
Area of rectangle HIJG =
= HI IJ
= (12) 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
(iii) In the figure, there are 5 squares, each of whose sides are 6 cm in length.
Area of the figure = 5 Area of square
= 5(side)2
= 5(6)2 cm2
= 180 cm2
Page No 231:
Answer:
(b) 28 cm
Let the length and the breadth of the rectangle be 7x cm and 5x cm, respectively.
It is given that the perimeter of the rectangle is 96 cm.
Perimeter of the rectangle = 2(7x+5x) cm
Page No 231:
Answer:
(d) 126 cm
Let length of the rectangle be L cm.
Area of the rectangle = 650 cm2
Area of the rectangle = () cm2
Perimeter of the rectangle = 2(Length + Breadth) cm = 2(50+13) cm = 126 cm
Page No 231:
Answer:
(b) Rs 2340
Perimeter of the rectangular field = 2(Length + Breadth)
= 2(34 + 18) m = 104 m
Cost of fencing 1 metre = Rs 22.50
Cost of fencing 104 m = Rs (22.50104) = Rs 2340
Page No 231:
Answer:
(b) 16 m
Total cost of fencing = Rs 2400
Rate of fencing = Rs 30/m
Perimeter of the rectangular field =
Let the breadth of the rectangular field be x m.
Perimeter of the rectangular field = 2(24 + x) m
Page No 231:
Answer:
(c) 17 m
Let the length and the breadth of the rectangle be L m and B m, respectively.
Area of the rectangular carpet = () m2
Perimeter of the rectangular carpet =
Page No 231:
Answer:
(a) 48 cm
Let the width and the length of the rectangle be x cm and 3x cm, respectively.
Applying Pythagoras theorem:
So, width of the rectangle is 6 cm.
Length of the rectangle =
Perimeter of the rectangle = 2(Length + Breadth) = 2(18 + 6) = 48 cm
Page No 231:
Answer:
(b) 2 : 1
Let the breadth of the plot be b cm.
Let the length of the plot be x cm.
Perimeter of the plot = 3x cm
Perimeter of the plot =2(Length + Breadth)= 2(x + b) cm
Page No 232:
Answer:
(b) 200 cm2
Area of the square =
Page No 232:
Answer:
(c) 20 m
Let one side of the square field be x m.
Total cost of fencing a square field = Rs 2000
Rate of fencing the field = Rs 25/m
Perimeter of the square field = () = 4x m
Page No 232:
Answer:
(b) 22 cm
Page No 232:
Answer:
(a) 28 cm
Page No 232:
Answer:
(b) 110 m
Page No 232:
Answer:
(d) 80000
Length of the road = 150 m = 15000 cm
Breadth of the road = 9 m = 900 cm
Area of the road = (Length × Breadth)
= (15000 × 900) cm2
= 13500000 cm2
Length of the brick = 22.5 cm
Breadth of the brick = 7.5 cm
Area of one brick = (Length × Breadth)
= ( 22.5 × 7.5 ) cm2
= 168.75 cm2
Page No 232:
Answer:
(b) 24.3 m2
Length of the room = 5 m 40 cm = 5.40 m
Breadth of the room = 4 m 50 cm = 4.50 m
Page No 232:
Answer:
(d) 16
Length of the sheet of paper = 72 cm
Breadth of the sheet of paper = 48 cm
Area of the sheet = (Length × Breadth)
⇒ ( 72 × 48 ) cm2 = 3456 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
Page No 233:
Answer:
(i) Sides of the triangle are 5.4 cm, 4.6 cm and 6.8 cm.
Perimeter of the triangle = (First side + Second side + Third Side)
= (5.4 + 4.6 + 6.8) cm = 16.8 cm
(ii) Length of each side of the given hexagon = 8 cm
∴ Perimeter of the hexagon = (6 × 8) cm = 48 cm
(iii) Length of the two equal sides = 6 cm
Length of the third side = 4.5 cm
∴ Perimeter of the triangle = {(2 × equal sides) + third side} cm = (2 × 6) + 4.5 = 16.5 cm
Page No 233:
Answer:
Let the length of the rectangle be x m.
Breadth of the rectangle = 75 m
Perimeter of the rectangle = 2(Length + Breadth)
= 2(x + 75) m = (2x + 150) m
It is given that the perimeter of the field is 360 m.
So, the length of the rectangle is 105 m.
Page No 233:
Answer:
Let the length of the rectangle be 5x m.
Breadth of the rectangle = 4x m
Perimeter of the rectangle = 2(Length + Breadth)
= 2(5x + 4x) m = 18x m
It is given that the perimeter of the field is 108 m.
∴ 18 x = 108
⇒ x =
∴ Length of the field = ( 5 × 6 )m = 30 m
Breadth of the field = ( 4 × 6 )m = 24 m
Page No 233:
Answer:
Let one side of the square be x cm.
Perimeter of the square =
It is given that the perimeter of the square is 84 cm.
Page No 233:
Answer:
Let the length of the room be x m.
Breadth of the room = 12 m
Area of the room = (Length × Breadth) = (x × 12) m2
It is given that the area of the room is 216 m2.
⇒ x × 12 = 216
⇒ x =
∴ Length of the rectangle = 18 m
Page No 233:
Answer:
Radius(r) of the given circle = 7 cm
Circumference of the circle, C = 2 πr
Hence, the circumference of the given circle is 44 cm.
Page No 233:
Answer:
Radius of the wheel =
⇒ r = cm
Circumference of the wheel = 2 πr
= cm
= 242 cm
In 1 revolution, the wheel covers a distance equal to its circumference.
∴ Distance covered by the wheel in 1 revolution = 242 cm
∴ Distance covered by the wheel in 500 revolutions = ( 500 × 242 ) cm
= 121000 cm (100 cm =1 m)
= 1210 m
Page No 233:
Answer:
Let the radius be r cm.
Diameter = 2 × Radius(r) = 2r cm
Circumference of the wheel = 2πr
∴ 2πr = 176
⇒ 2r = 56
Thus, the diameter of the given wheel is 56 cm.
Page No 233:
Answer:
Length of the rectangle = 36 cm
Breadth of the rectangle = 15 cm
Area of the rectangle = (Length × Breadth) sq units
= (36 × 15) cm2 = 540 cm2
Page No 233:
Answer:
(b) 64 cm
Side of the square = 16 cm
Perimeter of the square = (4 × side)
= (4 × 16) cm
= 64 cm
Page No 233:
Answer:
(a) 15 m
Let the breadth of the rectangle be x m.
Length of the rectangle = 16 m
Area of rectangle = (Length × Breadth) = (16 × x) m2
It is given that the area of the rectangle is 240 m2.
⇒ 16 × x = 240
⇒ x =
So, the breadth of the rectangle is 15 m.
Page No 233:
Answer:
(b) 225 m2
Page No 233:
Answer:
(a) 16 cm
Let one side of the square be x cm.
Area of the square = (Side )2 cm2 = x2 cm2
It is given that the area of the square is 256 cm2.
⇒ x2 = 256
⇒ x =
We know that the side of a square cannot be negative.
So, we will neglect −16.
Therefore, the side of the square is 16 cm.
Perimeter of the square =
Page No 233:
Answer:
(b) 10.5 m
Let the breadth of the rectangle be x m.
Length of the rectangle = 12 m
Area of the rectangle = 126 m2
Area of the rectangle =
It is given that the area of the rectangle is 126 m2.
Page No 233:
Answer:
(i) A polygon having all sides equal and all angles equal is called a regular polygon
(ii) Perimeter of a square = 4 × side
(iii) Area of a rectangle = (length) × (breadth)
(iv) Area of a square = (side)2
(v) If the length of a rectangle is 5 m and its breadth is 4 m, then its area is 20 m2
Area of a rectangle = (length) × (breadth) = (5×4) m2 = 20 m2
Page No 233:
Answer:
(a) Area of a rectangle | (iii) l × b |
(b) Area of a square | (iv) (side)2 |
(c) Perimeter of a rectangle | (v) 2(l + b) |
(d) Perimeter of a square | (ii) 4 × side |
(e) Area of a circle | (i) πr2 |
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