RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C.

Other Exercises

Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (b)
Length (l) = 12 cm
Breadth (b) = 9cm
height (h) = 8 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 1.1

Question 2.
Solution:
Total surface area of cube = 150 cm2
Side = \( \sqrt { \frac { 150 }{ 6 } } \)
= √25
= 5 cm
Volume = (side)3
= (5)3
= 125 cm3 (b)

Question 3.
Solution:
Volume of cube = 343 cm2
Side = \( \sqrt [ 3 ]{ 343 } =\sqrt [ 3 ]{ 7\times 7\times 7 } \)
= 7 cm
Total surface area = 6 (side)2
= 6 x (7)2
= 6 x 49 cm2
= 294 cm2 (c)

Question 4.
Solution:
Rate of painting = 10 paise per cm2
Total cost = Rs. 264.60
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 4.1

Question 5.
Solution:
Answer = (c)
Length of wall (l) = 8m = 800 cm
Breadth (b) = 22.5 cm
Height (h) = 6 m
= 600 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 5.1

Question 6.
Solution:
Answer = (c)
Edge of cube = 10 cm
Volume = a3 = (10)3 = 1000 cm3
Edge of box = 1 m = 100 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 6.1

Question 7.
Solution:
Answer = (a)
Ratio in sides of a cuboid = 1 : 2 : 3
Surface area = 88 cm2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 7.1

Question 8.
Solution:
Ratio in the two volumes = 1 : 27
Let volume of first volume = x3
and volume of second volume = 27x3
Side of first cube = x
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 8.1

Question 9.
Solution:
Surface area of a brick of measure 10 cm x 4 cm x 3 cm
= 2 (l x b + b x h + h x l)
= 2 [10 x 4 + 4 x 3 + 3 x 10] cm2
= 2 [40 + 12 + 30]
= 82 x 2
= 164 cm2 (c)

Question 10.
Solution:
Length of beam (l) = 9 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 10.1

Question 11.
Solution:
Water in rectangular reservoir = 42000
Volume = \(\\ \frac { 42000 }{ 1000 } \) = 42 m3
Length (l) = 6 m
Breadth (b) = 3.5 m
Depth = \(\\ \frac { volume }{ l\times b } \)
= \(\\ \frac { 42 }{ 6\times 3.5 } \)
= 2 m (c)

Question 12.
Solution:
Dimensions of a room are 10 m, 8 m, 3.3 m
Volume of air in it = lbh
= 10 x 8 x 3.3 = 264 m3
Air required for one man = 3 m3
No. of men = \(\\ \frac { 264 }{ 3 } \)
= 88 (b)

Question 13.
Solution:
Length of water tank (l) = 3 m
Width (b) = 2 m
and height (h) = 5 m
Volume = lbh = 3 x 2 x 5 = 30 m3
Water in it = 30 x 1000
= 30000 (a)

Question 14.
Solution:
Size of box = 25 cm, 15 cm, 8 cm
Surface area = (lb + bh + hl)
= 2 ( 25 x 15 + 15 x 8 + 8 x 25) cm2
= 2 (375 + 120 + 200) cm2
= 2(695)
= 1390 cm(b)

Question 15.
Solution:
Diagonal of cube = 4√3
Side = \( \frac { 4\sqrt { 3 } }{ \sqrt { 3 } } \)
= 4 cm
Volume = a3 = (4)3
= 64 cm3 (d)

Question 16.
Solution:
Diagonal of cube = 9√3 cm
Side = \( \frac { 9\sqrt { 3 } }{ \sqrt { 3 } } \)
= 9 cm
Surface area = 6a2
= 6 (9)2 = 6 x 81 cm2
= 486 cm2 (b)

Question 17.
Solution:
Let side of cube in first case = a
Then volume = a3
If side of cube is doubled, then side = 2a
Volume (2a)3 = 8a3
Becomes 8 times (d)

Question 18.
Solution:
Let side of cube in first case = a
Then surface area = 6a2
and side of second cube = 2a
Surface area = 6 (2a)2 = 6 x 4a2 = 24a2
Ratio = \(\frac { { 24a }^{ 2 } }{ { 6a }^{ 2 } } \) = 4
Becomes 4 times (b)

Question 19.
Solution:
Sides (edges) of 3 cubes are 6 cm, 8 cm, and 10 cm respectively
Volume of first cube = (6)3 = 216 cm3
Volume of second cube = (8)3 = 512 cm3
and volume of third cube
= (10)3 = 1000 cm3
Sum of volumes of 3 cubes = 216 + 512 + 1000
= 1728 cm3
Volume of new single cube = 1728 cm3
Edge = \(\sqrt [ 3 ]{ 1728 } \)
\(\sqrt [ 3 ]{ { \left( 12 \right) }^{ 3 } } \)
= 12 cm (a)

Question 20.
Solution:
Each edge of 5 cubes = 5 cm
Placing than adjacent to each other
Length of new cuboid (l)
= 5 x 5 = 25 cm
Breadth (b) = 5 cm
and height (h) = 5 cm
Volume of new cuboid = lbh
= 25 x 5 x 5 cm3
= 625 cm3 (d)

Question 21.
Solution:
Diameter of circular well = 2n
Radius = \(\\ \frac { 2 }{ 2 } \) = 1 m
Depth(h) = 14 m
Volume of earth dug out = πr2h
= \(\\ \frac { 22 }{ 7 } \) x 1 x 1 x 14
= 44 m (d)

Question 22.
Solution:
Capacity of cylindrical tank = 1848 m3
Diameter = 14 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 22.1

Question 23.
Solution:
Radius of a cylinder (r) = 20 cm
and height (h) = 60 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 23.1

Question 24.
Solution:
Radius of each coin (r) = 0.75 cm
and thickness (h) = 0.2 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 24.1

Question 25.
Solution:
Volume of silver = 66 cm3
Diameter of wire = 1 mm = \(\\ \frac { 1 }{ 10 } \)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 25.1

Question 26.
Solution:
Diameter of cylinder = 10 cm
Radius (r) = \(\\ \frac { 10 }{ 2 } \) = 5 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 26.1

Question 27.
Solution:
Diameter of cylinder = 7 cm
Radius (r) = \(\\ \frac { 7 }{ 2 } \) cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 27.1

Question 28.
Solution:
Curved surface area of a cylinder = 264 cm3
Height (h) = 14 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 28.1

Question 29.
Solution:
Diameter of cylinder = 14 cm
Radius (r) = 7 cm
Curved surface area = 220 cm2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 29.1

Question 30.
Solution:
Ratio in radii of two cylinder = 2 : 3
and ratio in their height = 5 : 3
Let radii of two cylinder = 2x and 3x
and corresponding heights = 5y, 3y
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 30.1

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B.

Other Exercises

Question 1.
Solution:
(i) Radius of the base of the cylinder (r) = 7 cm.
Height (h) = 50 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.3

Question 2.
Solution:
Radius of cylindrical tank (r) = 1.5 m
and height (h) = 10.5 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 2.1
= 74.25 x 1000l
= 74250 l

Question 3.
Solution:
Radius of the base of pole (r)
= 10 dm
= \(\\ \frac { 10 }{ 100 } \) m
= \(\\ \frac { 1 }{ 10 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 3.1

Question 4.
Solution:
Volume of cylinder = 1.54 m³
= 1540000 cm³
Diameter of its base = 140 cm
Radius (r) = 70 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 4.1

Question 5.
Solution:
Volume of cylindrical rod = 3850 cm³
Length of rod (h) = 1 m = 100 cm
Let radius of the base of the rod = r
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 5.1

Question 6.
Solution:
Diameter of closed cylinder = 14 m
Radius = \(\\ \frac { 14 }{ 2 } \)
= 7 m
Height = 5
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 6.1

Question 7.
Solution:
Circumference of the base of cylinder = 88 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 7.1

Question 8.
Solution:
Lateral surface of cylinder = 220 m²
Height (h) = 14 m
Let radius of cylinder = r
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 8.1

Question 9.
Solution:
Volume of cylinder = 1232 cm³
height (h) = 8cm
Let r be the radius, then
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 9.1

Question 10.
Solution:
Ratio in radius and height of a cylinder = 7 : 2
Let radius = 7x
then height = 2x
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 10.1

Question 11.
Solution:
Curved surface area = 4400 cm²
circumference of base = 110 cm
Let r be the radius
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 11.1

Question 12.
Solution:
In first case,
Side of square base (a) = 5 cm.
and height (h) = 14 cm.
Volume = 5 x 5 x 14 = 350 cm³
In second case,
Radius of the circular base (r) = 3.5 cm.
Height (h) = 12 cm.
Volume = πr²h
= \(\\ \frac { 22 }{ 7 } \) x 3.5 x 3.5 x 12 cm³
= 462 cm²
Hence second type of circular plastic can has greater capacity.
Difference = 462 – 350
= 112 cm³

Question 13.
Solution:
Diameter of a cylindrical pillar = 48 cm.
Radius (r) = \(\\ \frac { 48 }{ 2 } \) = 24 cm.
\(\\ \frac { 24 }{ 100 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 13.1

Question 14.
Solution:
Length of rectangular vessel (l) = 22 cm.
Breadth (A) = 16 cm.
and height (A) = 14 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 14.1

Question 15.
Solution:
Diameter of cylindrical metal = 1 cm.
Radius (r) = \(\\ \frac { 1 }{ 2 } \) cm.
Length. (A) = 11 cm.
Volume = πr²h
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 15.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 15.2

Question 16.
Solution:
Side of a solid cube = 2.2 cm
Volume = (side)³
= (2.2)³
= 10.648 cm³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 16.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 16.2

Question 17.
Solution:
Diameter of a well = 7 m
Radius (r) = \(\\ \frac { 7 }{ 2 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 17.1

Question 18.
Solution:
Inner diameter of well = 14 m
Inner radius = \(\\ \frac { 14 }{ 2 } \) = 7 m
Depth (h) = 12 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 18.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 18.2

Question 19.
Solution:
No. of revolutions = 750
Diameter of road roller = 84 cm
Length (h) = 1 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 19.1

Question 20.
Solution:
Thickness of the metal = 1.5 cm.
External diameter = 12 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 20.1

Question 21.
Solution:
Inner diameter of tube = 12 cm.
Inner radius (r) = \(\\ \frac { 12 }{ 2 } \) = 6 cm.
Thickness of metal = 1 m.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 21.1

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A

RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A.

Other Exercises

Question 1.
Solution:
(i)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.1
Length of cuboid (l) = 22 cm.
Breadth (b) = 12 cm
and height (h) = 7.5 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.3
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.4
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 1.5

Question 2.
Solution:
Length of water tank (l) = 2 m
75cm = 2.75 m
breadth (b) = 1 m 80cm = 1.80 m
and height (h) = 1 m 40 cm = 1.40 m
Volume of water filled in it = l.b.h = 2.75 x 1.80 x 1.40 m³
= 6.93 m³
Water in litres = 6.93 x 1000
= 6930 litres (1 m³ = 1000 litres) Ans.

Question 3.
Solution:
Length of iron (l) = 1.05 m
= 105 cm
breadth (b) = 70 cm and height (h) = 1.5 cm
volume of iron = l x b x h = 105 x 70 x 1.5 cm³
= 11025 cm³
weight of 1cm³ iron = 8 gram
Total weight = 11025 x g = 88200 g
= \(\\ \frac { 88200 }{ 1000 } \) kg
= 88.2 kg Ans.

Question 4.
Solution:
Area of courtyard = 3750 m²
Height of gravel = 1 cm.
Volume of gravel = 3750 x \(\\ \frac { 1 }{ 100 } \) m³
= 37.50 m³
Cost of 1 m³ gravel = Rs. 6.40
Total cost = Rs. 6.40 x 37.50
= Rs. 240 Ans.

Question 5.
Solution:
Length of hall (l) = 16 m
Breadth (b) = 12.5 m
height (h) = 4.5 m
Volume of air in it = l x b x h
= 16 x 12.5 x 4.5 m3
= 900 m³
Air for one person is required = 3.6 m³
Number of person which can be accommodated in the hall = 900 ÷ 3.6
= \(\\ \frac { 900\times 10 }{ 36 } \)
= 250

Question 6.
Solution:
Length of cardboard box (l)
= 1.2 m = 120 cm.
breadth (b) = 72 cm.
Height (h) = 54 cm.
Volume of box = l x b x h
= 120 x 72 x 54 cm³
= 466560 cm³
Volume of one soap bar = 6 x 4.5 x 4 cm³
= 108 cm³
No. of bars to be kept in it = \(\\ \frac { 466560 }{ 108 } \)
= 4320 Ans.

Question 7.
Solution:
Volume of one match box = 4 x 2.5 x 1.5 cm³ = 15 cm³
Volume of 144 matchboxes = 15 x 144 cm³
or volume of one packet = 2160 cm³
Length of carton (l) = 1.5 m = 150 cm
Breadth (b) = 84 cm
and height (h) = 60 cm.
Volume of one carton = l x b x h
= 150 x 84 x 60 cm³
= 756000 cm³
No. of packets = 756000 ÷ 2160
= 350 Ans.

Question 8.
Solution:
Length of one plank = 2m
= 200 cm
Breadth (b) = 25 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 8.1

Question 9.
Solution:
Length of wall (l) = 8m = 800 cm
Height (h) = 5.4 m = 540 cm
Width (b) = 33 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 9.1

Question 10.
Solution:
Length of wall (l) = 15 m
Width (b) = 30 cm = 0.3 m
Height (h) = 4 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 10.1

Question 11.
Solution:
Length of rectangular cistern (l) = 11.2 m
Breadth (b) = 6 m
Height (h) = 5.8 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 11.1

Question 12.
Solution:
Volume of block of gold = 0.5 m³
= 0.5 x 1000000 cm³
= 500000 cm³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 12.1

Question 13.
Solution:
Area of field = 2 hectare
= 20000 m²
Rainfall = 5 cm. = 0.05 m
Volume of water of rainfall
= Area of field x height of rainfall water
= 20000 x 0.05 m³
= 1000 m³ Ans.

Question 14.
Solution:
Speed of water = 3 km/h
Length of water flow in 1 minute
= \(\\ \frac { 3km }{ 60m } \)
= \(\\ \frac { 3000 }{ 60 } \)
= 50 m
Width of river = 45 m
Depth of river = 2 m
Volume of water in 1 minute
= 45 x 2 x 50 m³
= 4500 m³ Ans.

Question 15.
Solution:
Length of pit (l) = 5m
Width (b) = 3.5 m
Let depth of pit = h
then volume of earth dug out
= l.b.h = 5 x 3.5 x h = 17.5 h m³
But volume of earth = 14 m³
17.5 h = 14
h = \(\\ \frac { 14 }{ 17.5 } \) = \(\\ \frac { 140 }{ 175 } \)
=> h = \(\\ \frac { 4 }{ 5 } \) m
= \(\\ \frac { 4 }{ 5 } \) x 100
= 80 cm Ans.

Question 16.
Solution:
Width of tank = 90 cm = \(\\ \frac { 90 }{ 100 } \) m
Depth = 40 cm = \(\\ \frac { 40 }{ 100 } \) m
Water = 576 litre
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 16.1

Question 17.
Solution:
Volume of wood = 1.35 m³
Length of beam = 5m
Thickness = 36 cm = \(\\ \frac { 36 }{ 100 } \) m.
Width = \(\\ \frac { Volume }{ length\times thickness } \)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 7.1

Question 18.
Solution:
Volume of a room = 378 m³
Area of its floor = 84 m²
Height = \(\\ \frac { Volume }{ Area } \)
= \(\\ \frac { 378 }{ 84 } \) m
= 4.5 m Ans.

Question 19.
Solution:
Length of pool = 260 m
and width = 140 m.
Volume of water = 54600 m³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 19.1

Question 20.
Solution:
Outer length of wooden box (L) = 60 cm
Width (B) = 45 cm
and height (H) = 32 cm.
Thickness of wood used = 2.5 cm.
Inner length (l) = 60 – 2 x 2.5
= 60 – 5
= 55 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 20.1

Question 21.
Solution:
Outer length of open box = 36 cm
breadth = 25 cm
and height = 16.5 cm
thickness of iron = 1.5 cm.
∴ Inner length = 36 – 2 x 1.5
= 36 – 3
= 33 cm
breadth = 25 – 2 x 1.5
= 25 – 3
= 22 cm
and height = 16.5 – 1.5
= 15 cm .
∴ Volume of iron used in it = Outer volume – Inner volume
= 36 x 25 x 16.5 cm3 – 33 x 22 x 15 cm³
= 14850 – 10890
= 3960 cm³
weight of 1 cm³ = 8.5 gram
∴ Total weight = 3960 x 8.5 g
= 33660 g
= 33.660 kg
= 33.66 kg Ans.

Question 22.
Solution:
Outer length of the box = 56 cm
Width = 39 cm
and height = 30 cm
Volume = 56 x 39 x 30
= 65520 cm³
Thickness of wood used = 3cm.
∴ Inner length = 56 – 2 x 3
= 56 – 6
= 50 cm
Width = 39 – 2 x 3
= 39 – 6
= 33 cm
and height = 30 – 2 x 3
= 30 – 6
= 24 cm
∴ Inner volume of the box = 50 x 33 x 24 cm³
= 39600 cm³
and volume of wood used = Outer volume – Inner volume
= (65520 – 39600)cm³
= 25920 cm³ Ans.

Question 23.
Solution:
Outer length of box = 62 cm.
Outer width = 30 cm.
Outer height = 18 cm.
Thickness of wood = 2 cm.
∴ Internal length = 62 – 2 x 2
= 58 cm.
Internal width = 30 – 2 x 2
= 26 cm.
Internal height =18 – 2 x 2
= 14 cm.
Capacity of the box = lbh
= 58 x 26 x 14 cm³
= 21112 cm³ Ans.

Question 24.
Solution:
Outer length = 80 cm.
Outer width = 65 cm.
Outer height = 45 cm.
Total volume = 80 x 65 x 45 cm³
= 234000 cm³
Thickness of wood = 2.5 cm.
∴ Inner length = 80 – 2 x 2.5 = 75 cm.
Inner width = 65 – 2 x 2.5 = 60 cm.
Inner height = 45 – 2 x 2.5 = 40 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 24.1

Question 25.
Solution:
(i) Edge of cube (a) = 7 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.1
(a) Volume = a³ = (7)³
= 7 x 7 x 7 m³
= 343 m³ Ans.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 25.3

Question 26.
Solution:
Surface area of a cube = 1176 cm²
Let edge of the cube = a
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20A 26.1

Question 27.
Solution:
Volume of a cube = 729 cm³
Let edge of cube = a
then a³ = 729 = (9)³
a = 9 cm.
Hence surface area = 6a² = 6 (9)² cm²
= 6 x 81
= 486 cm² Ans.

Question 28.
Solution:
Length of metal block (l) = 2.25 m = 225 cm
Width (b) = 1.5 m = 150 cm
and height (h) = 27 cm
Volume of block = l x b x h
= 225 x 150 x 27 cm³
= 911250 cm³
Side of each cube = 45 cm.
Volume of each cube = a³
= 45 x 45 x 45
= 94125 cm³
Number of cubes = \(\\ \frac { 911250 }{ 91125 } \)
= 10 Ans.

Question 29.
Solution:
Let edge of given cube = a
Volume = a³
and surface area = 6a²
By doubling the edge of cube3 the side of new cube = a x 2 = 2a
Volume (2a)³ = 8a³
and surface area = 6 (2a)² = 6 x 4a²
= 24a² = 4 x 6a²
It is clear from the above that
Volume is increased 8 times and surface area is 4 times. Ans.

Question 30.
Solution:
Total cost of wood = Rs. 256
Rate = Rs.,500 per m³
Volume of wood = \(\\ \frac { 256 }{ 500 } \) = 0.512 m³
= 0.512 x 100 x 100 x 100 cm³
= 512000 cm³
Let length of each side = a
then a³ = 512000 = (80)³
a = 80
Hence length of each side = 80 cm. Ans.

 

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 19 Three-Dimensional Figures Ex 19B

RS Aggarwal Class 8 Solutions Chapter 19 Three-Dimensional Figures Ex 19B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 19 Three-Dimensional Figures Ex 19B.

Other Exercises

Question 1.
Solution:
Euler’s Relation
= F – E + V = 2
Where F is no. of faces
E is no. of edges
V is no. of vertices

Question 2.
Solution:
Edges of
(i) Cuboid are 12
(ii) Tetrahedron are 6
(iii) Triangular prism are 9
(iv) Square pyramid are – 8

Question 3.
Solution:
Faces of
(i) Cube are 6
(ii) Pentagonal prism are 7 (5 + 2)
(iii) Tetrahedron are 4
(iv) Pentagonal pyramid are 6

Question 4.
Solution:
Vertices of
(i) Cuboid are 8
(ii) Tetrahedron are 4
(iii) Pentagonal prism are 10
(iv) Square pyramid are 5

Question 5.
Solution:
(i) A cube
F – E + V = 2
=> 6 – 12 + 8 = 2
=> 14 – 12 = 2
=> 2 = 2
(ii) A tetrahedron
F – E + V = 2
=> 4 – 6 + 4 = 2
=> 8 – 6 = 2
=> 2 = 2
(iii) A triangular prism
F – E + V = 2
=> 5 – 9 + 6 = 2
=> 11 – 9
=> 2 = 2
(iv) A square pyramid
F – E + V = 2
=> 5 – 8 + 5 = 2
=> 10 – 8 = 2
=> 2 = 2

Hope given RS Aggarwal Solutions Class 8 Chapter 19 Three-Dimensional Figures Ex 19B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 19 Three-Dimensional Figures Ex 19A

RS Aggarwal Class 8 Solutions Chapter 19 Three-Dimensional Figures Ex 19A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 19 Three-Dimensional Figures Ex 19A.

Other Exercises

Question 1.
Solution:
(i) Cuboid : It has 6 faces.
(ii) Cube : It has 6 faces.
(iii) Triangular prism : It has 5 faces.
(iv) Square Pyramid : It has 5 faces.
(v) Tetrahedron : It has 4 faces. Ans.

Question 2.
Solution:
(i) Tetrahedron : It has 6 edges.
(ii) Rectangular pyramid : It has 8 edges
(iii) Cube : It is 12 edges.
(iv) Triangular prism : It has 9 edges Ans.

Question 3.
Solution:
(i) Cuboid : It has 8 vertices
(ii) Square pyramid : It has 5 vertices.
(iii) Tetrahedron : It is 4 vertices.
(iv) Triangular prism : It has 6 vertices

Question 4.
Solution:
(i) A cube has 8 vertices 12 edges and 6 faces.
(ii) The point at which three faces of a figure meet is known as vertex.
(iii) A cuboid is also known as a rectangular prism.
(iv) A triangular pyramid is called a tetrahedron.

 

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RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C

RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18C.

Other Exercises

Tick the correct answer in each of the following :

Question 1.
Solution:
Parallel sides 14 cm and 18 cm
Distance between parallel sides (h) = 9cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C 1.1

Question 2.
Solution:
Length of parallel sides are 19 cm and 13 cm
Area of trapezium = 128 cm²
Distance between then
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C 2.1

Question 3.
Solution:
Ratio in parallel sides = 3:4
Perpendicular distance (h) = 12 cm
Area of trapezium = 630 cm²
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C 3.1

Question 4.
Solution:
Area of trapezium = 180 cm²
and height (h) = 9 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C 4.1

Question 5.
Solution:
In the figure, AB || DC, DA ⊥ AB
DC = 7 cm, BC = 10 cm, AB = 13 cm
CL ⊥ AB
AD = DC = 7 cm
and LB – 13 – 7 = 6 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C 5.1
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18C 5.2

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RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B

RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18B.

Other Exercises

Question 1.
Solution:
In quad. ABCD
AC = 24 cm, BL ⊥ AC and DM ⊥ AC
BL = 8 cm and DM = 7 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 1.1

Question 2.
Solution:
In quad. ABCD, diagonal BD = 36 m
AL ⊥ BD and CM ⊥ BD
AL = 19 m and CM = 11 m
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 2.1

Question 3.
Solution:
In the given pentagon ABCDE,
BL ⊥ AC, DM ⊥ AC, EN ⊥ AC
AC = 18 cm, AM = 14 cm, AN = 6 cm,
BL = 4 cm, DM = 12 cm and EN = 9 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 3.1
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 3.2
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 3.3

Question 4.
Solution:
In hexagon ABCDEF, there are triangles and trapeziums
AP = 6 cm, PL = 2 cm, LN = 8 cm,
NM = 2 cm, MD = 3 cm, FP = 8 cm,
EN = 12 cm, BL = 8 cm and CM = 6 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 4.1
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 4.2
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 4.3
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 4.4

Question 5.
Solution:
In the given pentagon ABCDE,
BL ⊥ AC, CM ⊥ AD, EN ⊥ AD
AC = 10 cm, D = 12 cm, BL = 3 cm,
CM = 7 cm and EN = 5 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 5.1
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 5.2

Question 6.
Solution:
In the figure, ABCF is 0 square and CDEF is a trapezium
Now area of sq. ABCF
= (side)² = (20)² = 400 cm²
area of trap. CDEF
= \(\\ \frac { 1 }{ 2 } \) (ED + FC ) x height
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 6.1

Question 7.
Solution:
In the right ∆ABC
AB² = BC² + AC²
=> (5)² = (4)² + AC²
25 = 16 + AC²
AC² = 25 – 16 = 9 = (3)²
AC = 3 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 7.1
= 32 + 36
= 68 cm²

Question 8.
Solution:
AD = 23 cm, LM = 13 cm
AL = MD = \(\\ \frac { 23-13 }{ 2 } \) = \(\\ \frac { 10 }{ 2 } \) = 5 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18B 8.1

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RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A

RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 18 Area of a Trapezium and a Polygon Ex 18A.

Other Exercises

Question 1.
Solution:
In trapezium ABCD,
Length of parallel sides
AB = 24 cm, DC = 20 cm
and distance between them = 15 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 1.1

Question 2.
Solution:
Parallel sides of a trapezium ABCD are
l1 = 38.7 cm. and l2 = 22.3 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 2.1

Question 3.
Solution:
Parallel sides of the trapezium = 1 m, 1.4 m
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 3.1

Question 4.
Solution:
Area of trapezium = 1080 cm²
Lengths of parallel sides are
l1 = 55 cm and l2 = 35 cm
Let h be the distance between them
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 4.1

Question 5.
Solution:
Area of trapezium shaped field = 1586m²
Distance between parallel sides = 26 m
Sum of the parallel sides = \(\frac { Area\times 2 }{ Altitude }\)
= \(\frac { 1586\times 2 }{ 26 } \) = 122 m
One side = 84 m
Second side = 122 – 84
= 38 m

Question 6.
Solution:
Area of trapezium = 405 cm²
Ratio in parallel sides = 4:5
and distance between them = 18 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 6.1

Question 7.
Solution:
Area of trapezium = 180 cm²
Height (h) = 9 cm.
Let l1 and l2 be the parallel sides,
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 7.1

Question 8.
Solution:
Let one of parallel sides = x
Then second sides = 2x
Area = 9450 m²
Distance between them = 84 m
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 8.1

Question 9.
Solution:
Perimeter of trapezium ABCD = 130 m
AB ⊥ AD and BC
BC = 54 m, CD = 19 m, AD = 42 m
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 9.1

Question 10.
Solution:
In the given trapezium ABCD, AC is
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 10.1

Question 11.
Solution:
In trapezium ABCD,
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 11.1
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 11.2
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 11.3

Question 12.
Solution:
In trapezium ABCD,
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 12.1
AB || DC
AB = 25 cm DC = 1 cm
AD = 13 cm and BC = 15 cm
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 12.2
RS Aggarwal Class 8 Solutions Chapter 18 Area of a Trapezium and a Polygon Ex 18A 12.3

 

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RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B.

Other Exercises

Question 1.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 1.1
(i) Draw a line segment AB = 5.2 cm.
(ii) With centre A and radius 7.6 cm. and with centre B and radius 4.7 cm. draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 4.7 cm and with centre C and radius 5.2 cm draw arcs which intersect each other at D.
(v) Join AD and CD.
ABCD is the required parallelogram.

Question 2.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 4.3 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 2.1
(ii) With centre A and radius 4 cm. and with centre B and radius 6.8 cm., draw arcs which intersect each other at D.
(iii) Join AD and BD.
(iv) Again with centre B and radius 4 cm. and with centre D and radius 4.3 cm., draw arcs intersecting each other at C.
(v) Join DC and BC. ABCD is the required parallelogram.

Question 3.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At Q, draw a ray making an angle of 60° and cut off QR = 6 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 3.1
(iii) With centre P and radius 6 cm. and with centre R and radius 4 cm draw arcs intersecting each other at S.
(iv) Join RS and PS.
PQRS is the required parallelogram. Q.

Question 4.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 4.1
(ii) At C, draw a ray making an angle of 120° and cut off CD = 4.8 cm.
(iii) With centre B and radius 4.8 cm. with centre D and radius 5 cm, draw arcs intersecting each other at A.
(iv) Join AB and AD
ABCD is the required parallelogram.

Question 5.
Solution:
Steps of Construction :
We know that diagonals of a parallelogram bisect each other.
(i) Draw a line segment AB = 4.4 cm.
(ii) With centre A and radius \(\\ \frac { 5.6 }{ 2 } \) cm and with centre B and radius \(\\ \frac { 7 }{ 2 } \) = 3.5 cm. draw arcs
intersecting each other at O.
(iii) Join AO and BO and produce them to C and D respectively such that OC = 2.8 cm and OD = 3.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 5.1
(iv) Join AD, CD and BC
ABCD is the required parallelogram

Question 6.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 6.1
(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a perpendicular AX and cut off AL = 2.5 cm.
(iii) Through L, draw a line PQ parallel to AB.
(iv) From A, draw an arc of radius 3 -4 cm which intersects the line PQ at C.
(v) Join AC. BC
(vi) From PQ, cut off CD = AB.
(vii) Join AD
(viii) From C, draw a perpendicular CM to AB.
ABCD is the required parallelogram.

Question 7.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 7.1
(i) Draw a line segment AC = 3.8 cm.
(ii) Bisect it at O.
(iii) At O, draw a ray making an angle of 60° and produce it both sides.
(iv) From O cut off OB = OD = 2.3 cm.
(v) Join AB, BC, CD and AD.
ABCD is the required parallelogram.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 11 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 8.1
(ii) At B, draw a perpendicular and cut off BC = 8.5 cm.
(iii) With centre A and radius 8.5 cm and with centre C and radius 11 cm, draw arcs intersecting each other at D.
(iv) Join AD and CD.
ABCD is the required rectangle.

Question 9.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 9.1
(ii) At A and B draw perpendiculars and
cut off AD = BC = AB = 6.4 Cm.
(iii) Join CD.
ABCD is the required square.

Question 10.
Solution:
Steps of Construction :
(i) Draw a line segment AC = 5.8 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 10.1
(ii) Draw its perpendicular bisector intersecting AC at O.
(iii) From O, cut off OD = OB = 2.9 cm.
\(\qquad =\left( \frac { 1 }{ 2 } BD \right) \)
(iv) Join AB, BC, CD and DA. ABCD is the required square.

Question 11.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 11.1
(i) Draw a line segment QR = 3.6 cm.
(ii) At Q, draw a ray QX making an angle of 90°.
(iii) With centre R and radius 6 cm. draw an arc which intersects QX at P.
(iv) Join PR.
(v) With centre P and radius equal to QR and with centre R and radius equal to QP, draw arcs intersecting each other at S.
(vi) Join PS and RS.
PQRS is the required rectangle.
The length of other side PQ = 4.8 cm.

Question 12.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 12.1
(i)Draw a line segment AC = 8 cm.
(ii)Draw its perpendicular bisector intersecting it at O.
(iii)From O, cut off OB = OD = 3 cm.
(iv)Join AB, BC, CD and DA.
ABCD is the required rhombus.

Question 13.
Solution:
Steps of Construction :
(i)Draw a line segment AC = 6.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 13.1
(ii) With centres A and C and radius equal to 4 cm., draw arcs which intersect each other on both sides of line segment AC at B and D respectively.
(iii) Join AB, BC, CD and DA.
ABCD is the required rhombus.

Question 14.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 14.1
(i)Draw a line segmentAB = 7.2 cm.
(ii)At A draw a ray AX making an angle of 60° and cut off AD = 7.2 cm.
(iii)With centres D and B, and radius 7.2 cm., draw arcs intersecting each other at C.
(iv)Join CD and CB.
ABCD is the required rhombus.

Question 15.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 6 cm
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 15.1
(ii)At B, draw a ray BX making an angle of 75° and cut off BC = 4 cm.
(iii) At C, draw a ray CY making an angle of 180° – 75° = 105°
So that CY may be parallel to AB.
(iv) From CY, Cut off CD = 3.2 cm.
(v) Join DA.
ABCD is the required trapezium.

Question 16.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 7 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17B 16.1
(ii) At B, draw a ray BX making an angle of 60° and cut off BC = 5 cm.
(iii) At C, draw a ray CY making an angle of (180° – 60°) = 120° so that CY || AB.
(iv) With centre A and radius 6.5 cm. draw an arc intersecting CY at D.
(v) Join AD,
ABCD is the required trapezium.

Hope given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A

RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A.

Other Exercises

Question 1.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 4.2 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 1.1
(ii) With centre A and radius 8 cm and with centre B and radius 6 cm., draw arcs intersecting each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 5 cm. and with centre C, radius 5 2 cm. draw arcs intersecting each other at D.
(v) Join AD and CD. ABCD is the required quadrilateral.

Question 2.
Solution:
Steps of Construction :
(i) Draw a line segment PQ = 5.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 2.1
(ii) With Centre P and radius 4 cm. and with centre Q and radius 4.6 cm., draw arcs intersecting each other at R.
(iii) Join PR and QR.
(iv) Again with centre P and radius 3.5 cm. and with centre R and radius 4.3 cm. draw arcs intersecting each other at S.
(v) Join PS and RS. PQRS is the required quadrilateral.

Question 3.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 3.1
(ii) With centre A and radius 4.5 cm. and with centre B and radius 5.6 cm. draw arcs intersecting each other at D.
(iii) Join AD and BD.
(iv) With centre B and radius 3.8 cm. and with centre D and radius 4.5 cm., draw arcs intersecting each other at C.
(v) Join BC and DC. ABCD is the required quadrilateral.

Question 4.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.6 cm.
(ii) With centre A and radius 4.6 cm. and with centre B and radius 3.3 cm. draw arcs intersecting each other at C.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 4.1
(iii) Join AC and BC.
(iv) Again with centre A and radius 2.7 cm. and centre B and radius 4 cm., draw arcs intersecting each other at D.
(v) Join BD, AD and CD. ABCD is the required quadrilateral.

Question 5.
Solution:
Steps of Construction :
(i) Draw a line segment RS = 5 cm.
(ii) With centre R and S, radius 6 cm. each, draw arcs intersecting each other at R
(iii) Join PR and PS.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 5.1
(iv) With centre R and radius 7.5 cm. and with centre S and radius 10 cm, draw arcs intersecting each other at Q.
(v) Join RQ, SQ and PQ. PQRS is the required quadrilateral. Measuring the fourth sides PQ, it is 4.7 cm. (approx.)

Question 6.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 6.1
(ii) With centre A and radius 5.7 cm. and with centre B and radius 4 cm., draw arcs intersecting each other at D.
(iii) Join BD and AD.
(iv) Again with centre A and radius 8 cm and with centre D and radius 3 cm., draw arcs intersecting each other at C.
(v) Join AC, BC and DC. ABCD is the required quadrilateral

Question 7.
Solution:
Steps of Construction :
(i) Draw a line segment AB = 3.5 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 7.1
(ii) At B, draw a ray BX making an angle of 120° using protractor and cut off BC = 3.5 cm
(iii) With centres A and C and radius 5.2 cm, draw arcs intersecting each other at D.
(iv) Join CD and AD. ABCD is the required quadrilateral.

Question 8.
Solution:
Steps of Construction :
(i) Draw a line AB = 2.9 cm.
(ii) At A, draw a ray AX making an angle of 70° with AB. Using protractor and cut off AD = 3.4 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 8.1
(iii) With centre B and radius 3.2 cm and with centre D and radius 2.7 cm., draw arcs intersecting each other at C.
(iv) Join BC and DC. ABCD is the required quadrilateral.

Question 9.
Solution:
Steps of Construction
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 9.1
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray BX making an angle of 125° and cut off BA = 3.5 cm.
(iii) At C, draw a ray CY making an angle of 60° and cut off CD = 4.6 cm
(iv) Join AD. ABCD is the required quadrilateral.

Question 10.
Solution:
Steps of Construction :
(i) Draw a line segment QR = 5.6 cm.
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 10.1
(ii) At Q, draw a ray QX making an angle of 45° and cut off QP = 6 cm.
(iii) At R, draw a ray RY making an angle of 90° and cut off RS = 2.7 cm.
(iv) Join SP PQRS is the required quadrilateral.

Question 11.
Solution:
Steps of Construction :
∠A = 50°, ∠B = 105° and ∠D = 80°
and ∠A + ∠B + ∠C + ∠D = 360°
=> 50° + 105° + ∠C + 80° = 360°
=> ∠C + 235° = 360°
=> ∠C = 360° – 235°
=> ∠C = 125°
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 11.1
(i) Draw a line segment AB = 5.6 cm.
(ii) At B, draw a ray BY making an angle of 105° and cut off BC = 4 cm.
(iii) At C, draw a ray CZ making an. angle of 125° and at A, a ray AX making an angle of 50° intersecting each other at D.
then ∠D = 80°
ABCD is the required quadrilateral.

Question 12.
Solution:
∠P + ∠Q + ∠R + ∠S = 360°
100° + ∠Q + 100° + 75° = 360°
=> ∠Q + 275° = 360°
=> ∠Q = 360° – 275°
∠Q = 85°
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 12.1
(i) Draw a line segment PQ = 5 cm.
(it) At Q, draw a ray QX making an angle of 85° and cut off QR = 6.5 cm.
(iii) At R, draw a ray making an angle of 100° and at P, another ray making an angle of 100° which intersect each other at S. then ∠S = 75°
PQRS is the required quadrilateral.

Question 13.
Solution:
Steps of Construction :
RS Aggarwal Class 8 Solutions Chapter 17 Construction of Quadrilaterals Ex 17A 13.1
(i) Draw a line segment AB = 4 cm.
(ii) At B, draw a ray BX making an angle of 90°.
(iii) From A, draw an arc of 5 cm. radius intersecting BX at C.
(iv) Join AC.
(v) At C, draw a ray CY making an angle of 90°.
(vi) From A, draw an arc of radius 5.5 cm. which intersects CY at D.
(vii) Join AD.
ABCD is the required quadrilateral.

 

Hope given RS Aggarwal Solutions Class 8 Chapter 17 Construction of Quadrilaterals Ex 17A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B

RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 8. RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16B.

Other Exercises

Questions Tick the correct answer in each of the following.

Question 1.
Solution:
Answer = (c)
The diagonals of a rhombus are not necessarily equal but the diagonals in rectangle, square and isosceles trapezium are always equal.

Question 2.
Solution:
Answer = (c)
Each side of a rhombus
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B 2.1

Question 3.
Solution:
Answer = (b)
The sum of adjacent angles of a || gm = 180°
2x + 25° + 3x – 5° = 180°
=> 5x + 20° = 180°
=> 5x = 180° – 20° = 160°
=> x = \(\\ \frac { 160 }{ 5 } \)
= 32°

Question 4.
Solution:
Answer = (a)
The diagonals in rhombus, kite intersect each other at right angles.
But the diagonals of parallelogram do not necessarily intersect at right angles.

Question 5.
Solution:
Answer = (c)
Let l = 4x, b = 3x,
Then (diagonal)² = l² + b²
=> (25)² = 16x² + 9x²
=> 25x² = 625
=> x² = 25
=> x = 5
=> l = 4x = 4 x 5 = 20cm
b = 3x = 3 x 5 = 15cm
Perimeter = 2(l + b) = 2 (20 + 15)
= 2 x 35 = 70 cm

Question 6.
Solution:
Answer = (d)
AP and BP are the bisector of ∠A and ∠B
Sum of two adjacent angles of a ||gm = 180°
or ∠A + ∠B = 180°
RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16B 6.1
But ∠1 = \(\\ \frac { 1 }{ 2 } \) ∠A and ∠2 =\(\\ \frac { 1 }{ 2 } \) ∠B
∠1 + ∠2 = \(\\ \frac { 1 }{ 2 } \) ∠A + \(\\ \frac { 1 }{ 2 } \) ∠B
= \(\\ \frac { 1 }{ 2 } \) (∠A + ∠B)
= 180° x \(\\ \frac { 1 }{ 2 } \) = 90°
∠P = 180° – (∠1 + ∠2)
= 180° – 90° = 90°

Question 7.
Solution:
Answer = (b)
Let one adjacent angle = x
Then second angle (smallest) = \(\frac { 2 }{ 3 } x \)
x + \(\frac { 2 }{ 3 } x \) = 180°
= \(\frac { 5 }{ 3 } x \) = 180°
=> x = 180° x \(\\ \frac { 3 }{ 5 }\) = 108°
=> Smallest angle = 108° x \(\\ \frac { 2 }{ 3 }\) = 72°

Question 8.
Solution:
Answer = (a)
The diagonals of square, rhombus bisect the interior angle but the diagonals of a rectangle do not.

Question 9.
Solution:
Answer = (d)
Sides of a square are equal
2x + 3 = 3x – 5
=> 3x – 2x = 3 + 5
=> x = 8

Question 10.
Solution:
Answer = (c)
Let smallest angle = x
then largest angle = 2x – 24°
But x + 2x – 24° = 180°
=> 3x – 24° = 180°
=> 3x = 180° + 24 = 204°
=> x = \(\\ \frac { 204 }{ 3 }\) = 68°
largest angle = 180° – 68° = 112°

Hope given RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16B are helpful to complete your math homework.

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