Class 9 Maths RD Sharma Solutions

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS

Question 1.
If two cubes of side 6 cm are joined face to face, then find the volume of the resulting cuboid.
Solution:
Side of a cube = 6 cm
∴ By joining two such cubes, the length of so
formed cuboid (l) = 6 x 2 = 12 cm
Breadth (b) = 6 cm
Height (h) = 6 cm
∴ Volume = lbh = 12 x 6 x 6 cm³
= 432 cm³

Question 2.
Three cubes of metal whose edges are in the ratio 3 : 4 : 5, are melted down into a single cube whose diagonals is 12 \(\sqrt { 3 } \) cm. Find the edges of three cubes.
Solution:
Ratio in the sides of three cubes = 3 : 4 : 5
Let side of first cube = 3x
and side of second cube = 4x
and side of third cube = 5x
∴ Sum of volume of three cubes
= (3x)³ + (4x)³ + (5x)³
= 27x³ + 64x³ + 125x³ = 216x³
∴ Volume of the cube formed Jby melting these three cubes = 216x³

∴ Side of first cube = 3x = 3 x 2 = 6 cm
Side of second cube = 4x = 4×2 = 8 cm
and side of third cube = 5.r = 5 x 2 = 10 cm

Question 3.
If the perimeter of each face of a cube is 32 cm, find its lateral surface area. Note that four faces which meet the base of a cube are called its lateral faces.
Solution:
Perimeter of each face of a cube = 32 cm
∴ Length of edge = \(\frac { 32 }{ 4 }\) = 8 cm
and lateral surface area of the cube = 4 x (side)²
= 4 x 8 x 8 = 256 cm²

Question 4.
Find the edge of a cube whose surface area is 432 m².
Solution:

Question 5.
A cuboid has total surface area of 372 cm² and its lateral surface area is 180 cm², find the area of its base.
Solution:
Total surface area of a cuboid = 372 cm²
and lateral surface area = 180 cm²
∴ Area of base and roof = 372 – 180 = 192 cm²
and area of base = \(\frac { 192 }{ 2 }\) = 96 cm²

Question 6.
Three cubes of each side 4 cm are joined end to end. Find the surface area of the resulting cuboid.
Solution:
By joining three cubes of side 4 cm each, end is end, we get a cuboid
Length of cuboid = 4 x 3 = 12 cm
Breadth = 4 cm
and height = 4 cm

∴ Surface area = 2(lb + bh + hl)
= 2[12 x4+4×4 + 4x 12] cm²
= 2[48 + 16 + 48] cm²
= 2 x 112 = 224 cm²

Question 7.
The surface area of a cuboid is 1300 cm2. If its breadth is 10 cm and height is 20 cm, find its length.
Solution:
Surface area of a cuboid = 1300 cm²
Breadth (b) = 10 cm
and height (h) = 20 cm
Let l be the length, then
= 2 (lb + bh + hl) = 1300
lb+ bh + hl = \(\frac { 1300 }{ 2 }\) = 650
l x 10 + 10 x 20 + 20 x l = 650
10l + 20l + 200 = 650
⇒ 30l = 650 – 200 = 450
⇒ l = \(\frac { 450 }{ 30 }\) = 15
∴ Length of cuboid = 15 cm

Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS 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.

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS

Question 1.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? [NCERT]
Solution:
Length of water tank (l) = 6 m
Breadth (b) = 5 m
and depth (h) = 4.5 m

∴ Volume of water in it = lbh
= 6 x 5 x 4.5 m³ = 135 m³
Capacity of water in litres = 135 x 1000 litres (1 m³ = 1000 l)
= 135000 litres

Question 2.
A cubical vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? [NCERT]
Solution:
Length of vessal (l) = 10 m
Breadth (b) = 8 m
Volume = 380 m³

Question 3.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m³. [NCERT]
Solution:
Length of pit (l) = 8m
Width (b) = 6 m
and depth (h) = 3 m
∴ Volume of earth digout = lbh
= 8 x 6 x 3 = 144 m³
Cost of digging the pit at the rate of ₹30 per m³
= 144 x 30 = ₹4320

Question 4.
If the areas of three adjacent faces of a cuboid are 8 cm², 18 cm² and 25 cm². Find the volume of the cuboid.
Solution:
Let x, y, z be the three adjacent faces of the cuboid, then
x = 8 cm², y = 18 cm², z = 25 cm²
and let l, b, h are the dimensions of the cuboid, then
x = lb = 8 cm²
y = bh = 18 cm²
z = hl = 25 cm²
∴ Volume = lbh

Question 5.
The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.
Solution:
Let breadth of a room (b) = x

Question 6.
Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.
Solution:
Edge of first cube = 6 cm
Edge of second cube = 8 cm
and edge of third cube = 10 cm
∴ Volume of 3 cubes = (6)³ + (8)³ + (10)3 cm³
= 216 + 512 + 1000 cm³
= 1728 cm³
∴ Edge of so formed cube

Question 7.
Two cubes, each of volume 512 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each volume = 512 cm³
∴ Side (edge) = \(\sqrt [ 3 ]{ 512 }\)
=\(\sqrt [ 3 ]{ { 8 }^{ 3 } }\)  = 8 cm
Now by joining two cubes, then Length of so formed cuboid (l)
= 8 + 8 = 16 cm
Breadth (b) = 8 cm
and height (h) = 8 cm
∴ Surface area = 2(lb + bh + hl)
= 2[16 x 8 + 8 x 8 + 8 x 16] cm²
= 2[128 + 64 + 128] cm²
= 2 x 320 = 640 cm²

Question 8.
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
Solution:
Edge of metal cube = 12 cm
∴ Its volume = (Edge)³ = (12)³ cm3³
= 1728 cm³
It is melted and form 3 cubes
Edge of one smaller cube = 6 cm
and edge of second smaller cube = 8 cm
∴ Volume of two smaller cubes = (6)³ + (8)³ cm³
= 216 + 512 cm³ = 728 cm³
∴ Volume of third smaller cube = 1728 – 728 = 1000 cm³
∴ Edge of the third cube = \(\sqrt [ 3 ]{ 1000 }\)
= \(\sqrt [ 3 ]{ (10)3 }\)  cm = 10 cm

Question 9.
The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?
Solution:
Length of cinema hall (l) = 100 m
Breadth (b) = 50 m
and height (h) = 18 m
∴ Volume of air in it = lbh
= 100 x 50 x 18 m³ = 90000 m³
Air required for one person = 150 m³
∴ Number of persons in the hall = \(\frac { 90000 }{ 150 }\) = 600 persons

Question 10.
Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in width and 5 cm thick is 112 kg. Find the length of the block.
Solution:
Weight of 1 cm3 = 0.25 kg
Breadth of the block (b) = 28 cm
Thickness (h) = 5 cm
and total weight of the block = 112 kg

Question 11.
A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a liquid can be placed in it? Also, find the volume of the wood used in it.
Solution:
Outer length of the closed wooden box (l) = 25 cm
Breadth (b) = 18 cm
and height (h) = 15 cm
Width of wood = 2 cm

∴ Inner length = 25 – 2×2 = 25- 4 = 21cm
Breadth =18- 2×2 = 18-4 = 14 cm
and height =15- 2×2 = 15- 4=11 cm
Now outer volume = 25 x 18 x 15 cm³ = 6750 cm³
and inner volume = 21 x 14 x 11 cm³ = 3234 cm³
(i) Inner volume = 3234 cm³
(ii) Volume of wood = 6750 – 3234 = 3516 cm³

Question 12.
The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box?
Solution:
External length of a closed wooden box (L) = 48 cm
Width (B) = 36 cm
and height (H) = 30 cm
Thickness of wood = 1.5 cm

∴ Internal length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm
Width (b) = 36 – 2 x 1.5 cm
= 36 – 3 = 33 cm
and height (h) = 30 – 2 x 1.5 cm
= 30 – 3 = 27 cm
Now volume of internal box
= lbh = 45 x 33 x 27 cm³
Volume of one bricks = 6 x 3 x 0.75 cm³

Question 13.
A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.
Solution:
Edge of a cube = 9 cm
Volume of cube = (9)³ cm³
= 729 cm³
Now length of vessel (l) = 15 cm
and breadth (b) = 12 cm

Question 14.
A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7 m deep and the earth taken out is spread evenly on the field. By how many metres is the level of the field raised? Give the answer to the second place of decimal.
Solution:
Length of a field (l) = 200 m
Breadth (b) = 150 m
Length of plot = 50 m
and breadth = 40 m
Depth of plot = 7 m

Question 15.
A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a comer of the field and the earth taken out is spread- over the remaining area of the field. Find out the extent to which the level of the field has been raised.
Solution:
Length of a field (L) = 18m
and width (B) = 15 m
Length of pit (l) = 7.5 m
Breadth (b) = 6 m
and depth (h) = 0.8 m

∴ Volume of earth dugout = lbh
= 7.5 x 6 x 0.8 m³
= 45 x 0.8 = \(\frac { 45 x 4 }{ 5 }\) = 36 m3
Total area of the field = L x B
= 18 x 15 = 270 m²
and area of pit = lb = 7.5 x 6 = 45 m²
∴ Remaining area of the field excluding pit
= 270 – 45 = 225 m²
Let by spreading the earth on the remaining part of the field, the height = h
= 225 x h = 36
⇒ h = \(\frac { 36 }{ 225 }\) = \(\frac { 4 }{ 25 }\)= 0.16 m = 16 cm
∴ Level of field raised = 16 cm

Question 16.
A village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m x 15m x 6 m. For how many days will the water of this tank last? [NCERT]
Solution:
Population of a village = 4000
Water required per head per day = 150 litres
∴ Total water required = 4000 x 150 litres = 600000 litres
Dimensions of a tank = 20mx 15mx6m
∴ Volume of tank = 20 x 15 x 6 m3 = 1800 m³
Capacity of water in litres = 1800 x 1000 litres (1 m³ = 1000 litres)
= 1800000 litres
The water will last for = \(\frac { 1800000 }{ 600000 }\) = 3 days

Question 17.
A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the figure. If the edge of each cube is 3 cm, find the volume of the structure built by the child. [NCERT]

Solution:
No. of cubes at the given structure = 1+ 2 + 3+ 4 + 5 = 15
Edge of one cube = 3 cm
∴ Volume of one cube = (3)³ = 3 x 3 x 3 cm³ = 27 cm³
∴Volume of the structure = 27 x 15 cm³ = 405 cm³

Question 18.
A godown measures 40 m x 25 m x 10 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown. [NCERT]
Solution:
Length of godown (L) = 40 m
Breadth (B) = 25 m
and height (H) = 10 m
∴ Volume of godown = LBH
= 40 x 25 x 10 = 10000 m³
Dimension of one wooden crates = 1.5 m x 1.25 m x 0.5 m
∴Volume of one crate = 1.5 x 1.25 x 0.5 m³ = 0.9375 m³
∴ Number of crates to be stored in the

Question 19.
A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required? [NCERT]
Solution:
Length of wall (L) = 10 m = 1000 cm
Height (H) = 4 m = 400 cm
Thickness (B) = 24 cm = 24 cm
∴ Volume of wall = LBH = 1000 x 24 x 400 cm3 = 9600000 cm³
Dimensions of one brick = 24 cm x 12 cm x 8 cm = 2304 cm³

Question 20.
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that

Solution:
a, b, c are the dimensions of a cuboid S is the surface area and V is the volume
∴ V = abc and S = 2(lb + bc + ca)

Question 21.
The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, prove that V² = xyz.
Solution:
Let a, b, c are the dimensions of a cuboid then,
x = ab, y = bc, z = ca
and V = abc
Now L.H.S. = V²
= (abc)² = a²b²c²
= ab.bc.ca = xyz = R.H.S.
Hence V² = xyz

Question 22.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? [NCERT]
Solution:
Speed of water in a river = 2 km/hr

Question 23.
Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km per hour. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?
Solution:
Width of canal (b) = 30 dm = 3 m
Depth (h) = 12 dm = 1.2 m
Speed of water = 100 km/hr
Length of water flow in 30 minutes = \(\frac { 1 }{ 2 }\) hr

Question 24.
Half cubic metre of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold- sheet.
Solution:

Question 25.
How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 g, find the weight of the empty box in kg.
Solution:
External length of open box (L) = 36 cm
Breadth (B) = 25 cm
and Height (H) = 16.5 cm
Width of iron sheet used = 1.5 cm
∴ Inner length (l) = 36 – 1.5 x 2 = 36 – 3 = 33 cm
Breadth (b) = 25 – 2 x 1.5 = 25 – 3 = 22 cm
and Height (h) = 16.5 – 1.5 = 15 cm
∴ Volume of the iron used = Outer volume – Inner volume
= 36 x 25 x 16.5 – 33 x 22 x 15
= 14850 – 10890 = 3960 cm3
Weight of 1 cm3 = 15 g

Question 26.
A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.
Solution:
Base of the container = 5 cm x 5 cm
Level of water upto 1 cm from the top After placing a cube in it, the water rises to the top and 2 cubic cm of water overflows,
(i) ∴ Volume of water = 5 x 5 x 1 + 2 = 25 + 2 = 27 cm³
∴ Volume of cube = 27 cm³

Question 27.
A rectangular tank is 80 m long and 25 m broad. Water-flows into it through a pipe whose cross-section is 25 cm², at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes.
Solution:
Length of tank (l) = 80 m
Breadth (b) = 25 m
Area of cross section of the month of pipe = 25 cm²
and speed of water-flow =16 km/h
∴ Volume of water is 45 minutes

Question 28.
Water in a rectangular reservoir having base 80 m by 60 m is 6.5 m deep. In what time can the water be emptied by a pipe of which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 15 km/hr.
Solution:
Length of reservoir (l) = 80 m
Breadth (b) = 60 m
and depth (h) = 6.5 m
∴ Volume of water in it = lbh = 80 x 60 x 6.5 m³ = 31200 m³
Area of cross-section of the month of pipe = 20 x 20 = 400 cm²

Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2 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.

RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.2
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube VSAQS
• RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube MCQS

Question 1.
Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.
Solution:
Length of cuboid (l) = 80 cm
Breadth (b) = 40 cm
Height (h) = 20 cm
(i) ∴ Lateral surface area = 2h(l + b)
= 2 x 20(80 + 40) cm²
= 40 x 120 = 4800 cm²
(ii) Total surface area = 2(lb + bh + hl)
= 2(80 x 40 + 40 x 20 + 20 x 80) cm²
= 2(3200 + 800 + 1600) cm²
= 5600 x 2 = 11200 cm²

Question 2.
Find the lateral surface area and total surface area of a cube of edge 10 cm.
Solution:
Edge of cube (a) = 10 cm
(i) ∴ Lateral surface area = 4a²
= 4 x (10)² = 4 x 100 cm²= 400 cm²
(ii) Total surface area = 6a² = 6 x(10)² cm²
= 6 x 100 = 600 cm²

Question 3.
Find the ratio of the total surface area and lateral surface area of a cube.
Solution:
Let a be the edge of the cube, then Total surface area = 6a2²
and lateral surface area = 4a²
Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2

Question 4.
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?   [NCERT]
Solution:
Length of box (l) = 80 cm
Breadth (b) = 40 cm
and height (h) = 20 cm
∴ Total surface area = 2(lb + bh + hl)
= 2[80 x 40 + 40 x 20 + 20 x 80] cm²
= 2[3200 + 800 + 1600] cm² = 2 x 5600 = 11200 cm²
Size of paper sheet = 40 cm
∴ Area of one sheet = (40 cm)² = 1600 cm²
∴ No. of sheets required for the box = 11200 = 1600 = 7 sheets

Question 5.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m².
Solution:
Length of a room (l) = 5m
Breadth (b) = 4 m
and height (h) = 3 m
∴ Area of 4 walls = 2(l + b) x h
= 2(5 + 4) x 3 = 6 x 9 = 54 m²
and area of ceiling = l x b = 5 x 4 = 20 m²
∴ Total area = 54 + 20 = 74 m2
Rate of white washing = 7.50 per m²
∴ Total cost = ₹74 x 7.50 = ₹555

Question 6.
Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let each side of a cube = a cm
Then surface area = 6a² cm²
and surface area of 3 such cubes = 3 x 6a² = 18a² cm²
By placing three cubes side by side we get a cuboid whose,
Length (l) = a x 3 = 3a
Breadth (b) = a
Height (h) = a
∴ Total surface area = 2(lb + bh + hf)
= 2[3a x a+a x a+a x 3a] cm²
= 2[3a² + a² + 3a²] = 14 a²
∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

Question 7.
A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.
Solution:
Side of cube = 4 cm
But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64
Now surface area of one cube = 6 x (1)²
= 6 x 1=6 cm²
and surface area of 64 cubes = 6 x 64 cm² = 384 cm²

Question 8.
The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.
Solution:
Let h be the height of the room
Length (l) = 18 m
and width (b) = 12 m
Now surface area of floor and roof = 2 x lb = 2 x 18 x 12 m²
= 432 m²
and surface area of 4-walls = 2h (l + b)
= 2h(18 + 12) = 2 x 30h m² = 60h m²
∵ The surface are of 4-walls and area of floor and roof are equal
∴ 60h = 432
⇒ h = \(\frac { 432 }{ 60 }\) = \(\frac { 72 }{ 10 }\) m
∴ Height = 7.2m

Question 9.
Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if the cost of tiles is ₹360 per dozen. [NCERT]
Solution:
Edge of cubical tank = 1.5 m
∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m²
Area of floor = (1.5)² = 2.25 m²
∴ Total surface area = 9 + 2.25 = 11.25 m²
Edge of square tile = 25 m = 0.25 m²
∴ Area of 1 tile = (0.25)2 = .0625 m²

Question 10.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let edge of a cube = a
Total surface area = 6a2
By increasing edge at 50%,

Question 11.
A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of ₹5 per metre sheet, sheet being 2 m wide.
Solution:
Length of iron tank (l) = 12 m
Breadth (b) = 9 m
Depth (h) = 4 cm

Question 12.
Ravish wanted to make a temporary shelter for his car by making a box-like structure with tarpaulin that covers all the four sides and the top of tire car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make (he shelter of height 2.5 m with base dimensions 4 m x 3 m? [NCERT]
Solution:
Length of base (l) = 4m
Breadth (b) = 3 m
Height (h) = 2.5 m

Question 13.
An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of ₹50 per sq. metre.
Solution:
Length of open wood box (L) = 1.48 m = 148 cm
Breadth (B) = 1.16 m = 116 cm
and height (H) = 8.3 dm = 83 cm
Thickness of wood = 3 cm

Question 14.
The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at ₹3.50 per square metre.
Solution:
Length of room (l) = 12.5 m
Breadth (b) = 9 m
and height (h) = 7 m
∴ Total area of walls = 2h(l + b)
= 2 x 7[12.5 + 9] = 14 x 21.5 m² = 301 m²
Area of 2 doors of 2.5 m x 1.2 m
= 2 x 2.5 x 1.2 m² = 6 m²
and area of 4 window of 1.5 m x 1 m
= 4 x 1.5 x 1 = 6 m²
∴ Remaining area of walls = 301 – (6 + 6)
= 301 – 12 = 289 m²
Rate of painting the walls = ₹3.50 per m²
∴ Total cost = 289 x 3.50 = ₹1011.50

Question 15.
The paint in a certain container is sufficient to paint on area equal to 9.375 m2. How many bricks of dimension 22.5 cm x 10 cm x 7.5 cm can be painted out of this container? [NCERT]
Solution:
Area of place for painting = 9.375 m²
Dimension of one brick = 22.5 cm x 10 cm x 7.5 cm
∴ Surface area of one bricks = 2 (lb + bh + hl)
= 2[22.5 x 10 + 10 x 7.5 + 7.5 x 22.5] cm2
= 2[225 + 75 + 168.75]
= 2 x 468.75 cm² = 937.5 cm²

Question 16.
The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of ₹8 and ₹9.50 per m2 is ₹1248. Find the dimensions of the box.
Solution:
Ratio in the dimensions of a cuboidal box = 2 : 3 : 4
Let length (l) = 4x
Breadth (b) = 3.v
and height (h) = 2x
∴ Total surface area = 2 [lb + bh + hl]

Question 17.
The cost of preparing the walls of a room 12 m long at the rate of ₹1.35 per square metre is ₹340.20 and the cost of matting the floor at 85 paise per square metre is ₹91.80. Find the height of the room.
Solution:
Cost of preparing walls of a room = ₹340.20

Question 18.
The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 metres. The cost of decorating its walls (including doors and windows) at ₹6.60 per square metre is ₹5082. Find the length and breadth of the room
Solution:
Ratio in length and breadth = 4:3
and height (h) = 5.5 m
Cost of decorating the walls of a room including doors and windows = ₹5082
Rate = ₹6.60 per m²

Question 19.
A wooden bookshelf has external dimensions as follows: Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf.  [NCERT]
Solution:
Length (l) = 85 cm
Breadth (b) = 25 cm
and height (h) = 110 cm
Thickness of plank = 5 cm
Surface area to be polished
= [(100 x 85) + 2 (110 x 25) + 2 (85 x 25) + 2 (110 x 5) + 4 (75 x 5)]
= (9350 + 5500 + 4250 + 1100 + 1500) cm² = 21700 cm²

Hope given RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume of a Cuboid and Cube Ex 18.1 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.

RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
• RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
• RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

Mark the correct alternative in each of the following:
Question 1.
The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is
(a) 225 cm²
(b) 225\(\sqrt { 3 } \) cm²
(c) 225\(\sqrt { 2 } \) cm²
(d) 240 cm²
Solution:
Sides of triangle and 16 cm, 30 cm, 34 cm

Question 2.
The base of an isosceles right triangle is 30 cm. Its area is
(a) 225 cm²
(b) 225\(\sqrt { 3 } \) cm²
(c) 225\(\sqrt { 2 } \) cm²
(d) 450 cm²
Solution:
Base of isosceles triangle ∆ABC = 30cm
Let each of equal sides = x
Then AB = AC = x
Now in right ∆ABC,

Question 3.
The sides of a triangle are 7cm, 9cm and 14cm. Its area is
(a) 12\(\sqrt { 5 } \) cm²
(b) 12\(\sqrt { 3 } \) cm²
(c) 24 \(\sqrt { 5 } \) cm²
(d) 63 cm²
Solution:

Question 4.
The sides of a triangular field are 325 m, 300 m and 125 m. Its area is
(a) 18750 m²
(b) 37500 m²
(c) 97500 m²
(d) 48750 m²
Solution:
Sides of a triangular field are 325m, 300m, 125m

Question 5.
The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is
(a) 20 cm
(b) 30 cm
(c) 40 cm
(d) 50 cm
Solution:
The sides of a triangle are 50 cm, 78 cm, 112cm

Question 6.
The sides of a triangle are 11m, 60m and 61m. Altitude to the smallest side is
(a) 11m
(b) 66 m
(c) 50 m
(d) 60 m
Solution:
Sides of a triangle are 11m, 60m and 61m

Question 7.
The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

Solution:
Sides of a triangle are 11 cm, 15 cm, 16 cm

Question 8.
The base and hypotenuse of a right triangle are respectively 5cm and 13cm long. Its area is
(a) 25 cm²
(b) 28 cm²
(c) 30 cm²
(d) 40 cm²
Solution:
In a right triangle base = 5 cm
base hypotenuse = 13 cm

Question 9.
The length of each side of an equilateral triangle of area 4 \(\sqrt { 3 } \) cm², is

Solution:
Area of an equilateral triangle = 4\(\sqrt { 3 } \) cm²
Let each side be = a

Question 10.
If the area of an isosceles right triangle is 8cm, what is the perimeter of the triangle.
(a) 8 + \(\sqrt { 2 } \) cm²
(b) 8 + 4\(\sqrt { 2 } \) cm²
(c) 4 + 8\(\sqrt { 2 } \) cm²
(b) 12\(\sqrt { 2 } \) cm²
Solution:
Let base = x
ABC an isosceles right triangle, which has 2 sides same
⇒ Height = x

Question 11.
The length of the sides of ∆ABC are consecutive integers. If ∆ABC has the same perimeter as an equilateral triangle with a side of length 9cm, what is the length of the shortest side of ∆ABC?
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
Side of an equilateral triangle = 9 cm
Its perimeter = 3 x 9 = 27 cm
Now perimeter of ∆ABC = 27 cm
and let its sides be x, x + 1, x +2

Question 12.
In the figure, the ratio of AD to DC is 3 to 2. If the area of ∆ABC is 40cm2, what is the area of ∆BDC?
(a) 16 cm²
(b) 24 cm²
(c) 30 cm²
(d) 36 cm²
Solution:
Ratio in AD : DC = 3:2
and area ∆ABC = 40 cm²

Question 13.
If the length of a median of an equilateral triangle is x cm, then its area is

Solution:
∵ The median of an equilateral triangle is the perpendicular to the base also,
∴ Let side of the triangle = a

Question 14.
If every side of a triangle is doubled, then increase in the area of the triangle is
(a) 100\(\sqrt { 2 } \) %
(b) 200%
(c) 300%
(d) 400%
Solution:
Let the sides of the original triangle be a, b, c

Question 15.
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 1272 cm, then area of the triangle is

Solution:
A square and an equilateral triangle have equal perimeter

Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS 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.

RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1
• RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.2
• RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS
• RD Sharma Class 9 Solutions Chapter 17 Constructions MCQS

Question 1.
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
In ∆ABC,
Base BC = 5cm
Altitude AD = 4cm

Question 2.
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
Sides of triangle are 3 cm, 4cm and 5cm

Question 3.
Find the area of an isosceles triangle having the base x cm and one side y cm.
Solution:
In isosceles ∆ABC,
AB = AC = y cm
BC = x cm

Question 4.
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Each side of equilateral triangle (a) = 4cm

Question 5.
Find the area of an equilateral triangle having each side x cm.
Solution:

Question 6.
The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Solution:
Perimeter of the field = 144 m
Ratio in the sides = 3:4:5
Sum of ratios = 3 + 4 + 5 = 12

Question 7.
Find the area of an equilateral triangle having altitude h cm.
Solution:
Altitude of an equilateral triangle = h
Let side of equilateral triangle = x

Question 8.
Let ∆ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
Solution:
Let a, b, c be the sides of the original triangle

Hence area of new triangle = 4 x area of original triangle.

Question 9.
If each side of a triangle is doubled, then find percentage increase in its area.
Solution:
Sides of original triangle be a, b, c

Question 10.
If each side of an equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
Solution:
Let the sides of the original triangle be a, b, c and area ∆, then

Hope given RD Sharma Class 9 Solutions Chapter 17 Constructions VSAQS 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.