A+ Guide to It Technical Support 9th Edition Ch 18 Review Answers

RD Sharma Solutions Class ix Maths Chapter 18 – Gratuitous PDF Download

RD Sharma Solutions for Class 9 Maths Chapter eighteen Surface Area and Book of Cuboid and Cube consists of various topics which are of import from an exam perspective. It is essential to follow the right source for effective learning of concepts. Hence, in order to help students with the right report material, experts have prepared the solutions to all the textbook questions in a precise fashion. Students who practice these solutions on a regular basis understand the concepts with ease. The master objective of developing RD Sharma Solutions for Class nine is to help students ace the exam more effectively.

Our RD Sharma Solutions for Class 9 are designed according to the 2021-22 syllabus prescribed by the Central Board of Secondary Educational activity. Here students will find three exercises of Class 9 Maths Chapter 18 Surface Area and Volume of Cuboid and Cube . Click on the below link to access all chapter 18 exercises.

Affiliate xviii Surface Expanse And Volume…

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Admission Answers to Maths RD Sharma Solutions for Course 9 Chapter eighteen Surface Area and Volume of Cuboid and Cube

Practise 18.ane Folio No: 18.14

Question i: Find the lateral surface surface area and total surface expanse of a cuboid of length lxxx cm, latitude 40 cm and pinnacle 20 cm.

Solution:

Given, Dimensions of a cuboid:

Length (50) = 80 cm

Breadth (b) = 40 cm

Superlative (h) = twenty cm

We know that, Full Surface Area = 2[lb + bh + hl]

By substituting the values, we go

= ii[(80)(twoscore)+(40)(xx)+(20)(lxxx)]

= ii[3200+800+1600]

= 2[5600]

= 11200

Therefore, Total Surface Area = 11200 cm²

At present,

Lateral Surface Area = 2[l + b]h

= two[eighty+40]20

= 40[120]

= 4800

Thus, Lateral Surface area is 4800 cm^two.

Question two: Discover the lateral surface expanse and full surface expanse of a cube of edge 10 cm.

Solution:

Side of a Cube = 10 cm (Given)

Formula for Cube Lateral Surface Area = iv side^two

Cube Lateral Surface Expanse = four(10 x 10)

= 400 cm²

Full Surface Area = 6 Sideⁱⁱ

= 6(10²)

= 600 cmⁱⁱ

Question 3: Find the ratio of the total expanse and lateral area of a cube.

Solution:

Total Surface Surface area of the Cube (TSA) = 6 Side²

Lateral surface area of the Cube (LSA) = 4 Side²

Now,

Ratio of TSA and LSA = (6 Side²)/ (iv Sideⁱⁱ) = 3/2 or three:2.

Question 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the verbal quantity of paper to buy for this purpose. If the box has length, breadth, and superlative every bit 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she crave?

Solution:

The dimensions of the wooden block are:

Length (l) = 80cm

Breadth (b) = 40cm

Height (h) = 20cm

Surface Area of the wooden box = 2[lb + bh + hl]

= 2[(eighty×40)+(40×20)+(twenty×80)]

= ii[5600]

= 11200

Surface Area of the wooden box is 11200 cmⁱⁱ

The Expanse of each canvas of the paper = twoscore×40 cm^two = 1600 cm²

At present,

The total number of sheets required = (Surface area of the box )/(Area of one sheet of newspaper)

= 11200/1600

= 7

Therefore, Marry would crave seven sheets.

Question 5: The length, latitude, and height of a room are 5 m, 4 chiliad and 3 g respectively. Notice the toll of white washing the walls of the room and the ceiling at the charge per unit of Rs 7.50 m².

Solution:

Formula: Total Area to be done = lb + 2(l + b)h ….(ane)

Where, l = length , b = breadth and h = height.

From given:

Length = l = v m

Breadth = b = 4 m

Height = h = 3 m

Total area to exist white washed = (5×iv) + two(5 + 4)iii

(using (ane))

= 74

Total surface area to be white done is 74 m²

Now, cost of white washing one yard² is Rs. 7.50 (Given)

Therefore, the price of white washing 74 m² = (74 x 7.50)

= Rs. 555

Question 6: Three equal cubes are placed adjacently in a row. Find the ratio of a total expanse of the new cuboid to that of the sum of the surface areas of the three cubes.

Solution:

Let latitude of the cuboid = a

And then, length of the new cuboid = 3a and

Height of the new cuboid = a

Now,

Total area of the new cuboid (TSA) = 2(lb+bh+hl)

= 2(3a 10 a + a x a + a x 3a)

= 14a^two

Over again,

Total Surface area of 3 cubes = iii x (half-dozen side² )

= 3 x 6a²

= 18aⁱⁱ

Therefore, ratio of a full surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a^two/18a^two = 7/9 or vii:9

Therefore, required ratio is 7:nine. Answer.

Question 7: A 4 cm cube is cut into one cm cubes. Calculate the total surface area of all the small cubes.

Solution:

Edge of the cube = 4 cm (Given)

Volume of the cube = Side³ = 4³ = 64

Volume of the cube is 64cm³

Once again,

Edge of the cube = ane cm³

So, Total number of small-scale cubes = 64cm^three/1cm³ = 64

And, total surface area of all the cubes = 64 x half-dozen x 1 = 384 cmⁱⁱ

Question 8: The length of a hall is 18 m and the width 12 m. The sum of the areas of the flooring and the flat roof is equal to the sum of the areas of the iv walls. Find the summit of the hall.

Solution:

Dimensions of the hall are:

Length = 18m

Width = 12m

From statement:

Area of the flooring and the flat roof = Sum of the areas of four walls …(1)

Using respective formulas and given values, nosotros have

Area of the flooring and the apartment roof = 2lb = 2 x 18 10 12 = 432 sq/ft …(2)

Sum of the areas of four walls = (ii 10 18h + 2 x 12h)sq/ft …(3)

Using equation (2) and (iii) in (1), we get

432 = 2 ten 18h + 2x12h

18h + 12h = 216

or h = 7.two

Therefore, height of the hall is 7.ii g.

Question 9: Hameed has built a cubical water tank with lid for his firm, with each other edge 1.five m long. He gets the outer surface of the tank excluding the base, covered with foursquare tiles of side 25 cm. Observe how much he would spend for the tiles if the price of tiles is Rs 360 per dozen.

Solution:

Edge of the cubical tank = ane.5m or 150 cm

Surface surface area of the cubical tank (5 faces) = 5 x Area of 1 Confront

= 5 x (150 x 150) cmⁱⁱ …….(1)

Find area of each square tile:

Side of tile = 25 cm (given)

Surface area of 1 tile = 25 x 25 cm² …….(two)

At present,

Number of tiles required = (Surface Area of Tank ) / (Expanse of each Tile)

= (5×150×150) / 25×25

= 180

Find cost of tiles:

Toll of ane dozen tiles, i.eastward., toll of 12 tiles = Rs. 360

Therefore, cost of one tile = Rs.360/12 = Rs.30

Then, the cost of 180 tiles = 180 x xxx = Rs. 5400

Question 10: Each edge of a cube is increased by 50%. Notice the percentage increase in the surface surface area of the cube.

Solution:

Let 'a' exist the border of a cube.

Surface area of the cube having border 'a' = 6aⁱⁱ …..(1)

As given, a new border after increasing existing edge past fifty%, we get

The new border = a + 50 a /100

= 3a/2

Surface area of the cube having border '3a/2' = half-dozen x (3a/2)² = (27/two) aⁱⁱ ……(2)

Decrease equation (i) from (2) to find the increment in the Surface Area:

Increase in the Expanse = (27/two) a^two – 6a²

= (15/2)a²

Now,

Per centum increase in the surface area = ( (fifteen/two)a² / 6a² ) x 100

= 15/12 x 100

= 125%

Therefore, percentage increase in the surface area of a cube is 125.

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Practice eighteen.2 Page No: 18.29

Question 1: A cuboidal water tank is 6 one thousand long, 5 m broad and 4.5 1000 deep. How many liters of water can it hold?

Solution:

Dimensions of a cuboidal water tank:

Length = l = 6m

Breadth = b = 5m

Height = h = 4.5m

We know, Volume of the cuboidal water tank = lbh

Past substituting the values, we get

Volume = 6×5×iv.5

= 135

Therefore, Volume of the cuboidal water tank is 135 yardⁱⁱⁱ

Convert into liters:

We know; 1 mⁱⁱⁱ = 1000 liters

So, 135m^three = (135×1000)liters

= 135000 liters

Hence, the tank can hold i,35,000 liters of water.

Question 2: A cuboidal vessel is x m long and 8 m wide. How loftier must information technology be made to hold 380 cubic meters of a liquid?

Solution:

Dimensions of a cuboidal vessel:

Length = l = 10 yard

Breadth = b = eight one thousand

Volume of the vessel = 380 g^three (given)

Let 'h' be the summit of the cuboidal vessel.

We know, Volume of cuboidal vessel = lbh

lbh = 380 m^three

or 10×8×h = 380

or h = (380)/(ten×8)

or h = 4.75

Therefore, height of the vessel should exist 4.75 yard.

Question 3: Observe the cost of digging a cuboidal pit viii m long, half dozen m wide and 3 yard deep at the rate of Rs 30 per m³.

Solution:

Dimensions of a cuboidal pit:

Length = l = 8 m

Breadth = b = 6 m

Depth or height = h = 3 m

We know, Volume of the Cuboidal pit = lbh

= eight×six×3

= 144

Volume of the Cuboidal pit is 144 m³

Now, discover the price:

Cost of digging one m³ = Rs. 30 (Given)

Cost of digging 144 chiliad³ = 144 x 30 = Rs. 4320

Question 4: If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then evidence that

Solution:

Dimensions of a cube are:

Length = l = a

Latitude = b = b

Peak = h = c

We know, Volume of the cube (5) = lbh

= a×b×c

Or V = abc ….(i)

Again,

Surface expanse of the cube (S) = 2 (lb+bh+hl)

or S = 2 (ab+bc+ca) ……(2)

Now,

Hence Proved.

Question v: The areas of 3 side by side faces of a cuboid are 10, y and z. If the volume is Five, Testify that V² = xyz.

Solution:

Permit a, b and c exist the length, breadth, and top of the cuboid.

Then, x = ab, y = bc and z = ca

[Since areas of iii side by side faces of a cuboid are ten, y and z (Given)]

And xyz = ab × bc × ca = (abc)² ……(one)

Nosotros know, Volume of a cuboid ( V ) = abc …..(2)

From equation (1) and (two), we have

V² = xyz

Hence proved.

Question half dozen: If the areas of 3 adjacent face of a cuboid are eight cm² , 18 cm² and 25 cm². Discover the volume of the cuboid.

Solution:

Let x, y, z denote the areas of iii next faces of a cuboid, then,

10 = 50×b = eight cm^two

y = b×h = 18 cm^two

z = fifty×h = 25 cmⁱⁱ

Where l = length of a cuboid, b = breadth of a cuboid and h = acme of a cuboid

xyz = viii × eighteen × 25 = 3600 ….(i)

Book of cuboid (V) = lbh

From above results, nosotros can write,

xyz = lb×bh×lh = (lbh)² = 5² . …..(ii)

Form equation (one) and (2), Nosotros get

Vⁱⁱ = 3600

or V = threescore

Thus, Book of the cuboid is 60 cmⁱⁱⁱ

Question 7: The breadth of a room is twice its summit, one half of its length and the volume of the room is 512 cu.dm. Observe its dimensions.

Solution:

Let, l, b and h are the length, breadth and height of the room.

As per given argument,

b = 2h and b = 50/2

⇒ l/2 = 2h

or l = 4h

At present, nosotros take l = 4h and b = 2h

We know, Volume of the room = lbh

Volume of the room = 512dm³ (given)

So, 4h×2h×h=512

or hⁱⁱⁱ=64

or h=four

Therefore, Length of the room (l) = 4h = 4×4 = 16 dm

Breadth of the room (b) = 2h = 2×4 = 8 dm

And Height of the room (h) = four dm.

Question 8: A river iii k deep and xl yard broad is flowing at the charge per unit of ii km per hour. How much h2o will autumn into the sea in a infinitesimal?

Solution:

Water menstruation of a river = 2 km per hour = (2000/60) m/min or (100/3 )m/min

[we know: 1 km = 1000 m and ane hour = lx mins]

Depth of the river (h) = 3m

Width of the river (b) = 40m

Volume of the h2o flowing in i min = 100/3 × 40 × 3 = 4000 g³

Or 4000 m³ = 4000000 litres

Therefore, in 1 infinitesimal 4000000 litres of water will autumn in the ocean.

Question ix: Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km every hour. What much area will it irrigate in 30 minutes if 8 cm of continuing h2o is desired?

Solution:

Water in the canal forms a cuboid of Width (b) and Height (h).

b = 30dm = 3m and h = 12dm = 1.2m

Here, Cuboid length = distance travelled in thirty min with a speed of 100 km per hour.

Therefore, Length of the cuboid (l) = 100 × xxx/60 = 50000 metres

Book of water used for irrigation = lbh = 50000×3×1.2 m³

Water accumulated in the field forms a cuboid of base area equal to the area of the field and height = 8/100 metres (Given)

Therefore, Area of field × 8/100 = 50000 × iii × 1.ii

Area of field = (50000 × 3 × i.2) × 100/8

= 2250000

Thus, expanse of field is 2250000 thousand² .

Question 10: Three metal cubes with edges 6cm, 8cm, 10cm respectively are melted together and formed into a single cube. Find the volume, area and diagonal of the new cube.

Solution:

Let u.s.a. consider, 'ten' exist the length of each edge of the new cube.

Volume of cube = 10³

⇒ xⁱⁱⁱ = (6³ + viiiⁱⁱⁱ + 10³)cm³

or ten³ = 1728

or ten=12

Volume of the new cube = x³ = 1728 cm³

Surface surface area of the new cube = half-dozen(side)² = 6(12)² = 864 cm²

And, diagonal of the newly formed cube = √3a = 12√3 cm

Question 11: Ii cubes, each of volume 512 cmⁱⁱⁱ are joined stop to stop. Find the surface area of the resulting cuboid.

Solution:

Allow 'a' be the side of a cube.

Volume of the cube = 512cm^three (Given)

We know volume cube = (side)^three

⇒ a^3 = 512

or a = eight

Each side of a cube is eight cm.

Now,

Dimensions of the new cuboid formed are:

Length (l) = 8+8 = sixteen cm,

Breadth (b) = 8 cm and

Height (h) = 8 cm

Surface area = ii(lb+bh+hl)

= 2 (16×eight+8×eight+16×8)

= 640 cmⁱⁱ

Therefore, Surface area of a cube is 640 cm².

Question 12: Half cubic meter of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold-sheet.

Solution:

Volume of gilded-canvass = 1/2 grand^three or 0.5 m³ (Given)

Area of the gold-sheet = one hectare i.eastward. 10000 m²

Thickness of golden sheet = (Book of solid)/(Area of gold sheet)

= 0.v m³/10000 mⁱⁱ

= m/20000

Or Thickness of aureate sail = 1/200 cm

[1 chiliad = 100 cm]

Therefore, thickness of the silver canvass is i/200 cm.

Question 13: A metal cube of edge 12 cm is melted and formed into 3 smaller cubes. If the edges of the 2 smaller cubes are 6 cm and 8 cm, observe the edge of the tertiary smaller cube.

Solution:

From the given statement, nosotros have

Volume of the big cube = v1 + v2 + v3

Permit the edge of the tertiary cube be 'x' cm

12³ = 6^three + 8³ + x^three

[Using formula, Volume of cube = (side)³]

1728 = 216 + 512 + x³

or g = x³

or x = x

Therefore, length of the third side is 10 cm.

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Exercise VSAQs Page No: 18.35

Question 1: If two cubes each of side six cm are joined confront to face, then find the volume of the resulting cuboid.

Solution:

Side of ii equal cubes = half dozen cm (Given)

When we join, two cubes face to face up formed a cuboid.

Dimensions of a cuboid are:

Length = half dozen cm + 6 cm = 12 cm

Breadth = 6 cm

Meridian = 6 cm

Therefore, volume of cuboid = lbh = 12 × 6 × 6 = 432 cm³

Question 2: Iii cubes of metal whose edges are in the ratio iii : iv : 5 are melted down into a single cube whose diagonal is 12√3 cm. Notice the edges of three cubes.

Solution:

Given:

Ratio of edge of 3 cubes = 3 : 4 : 5

Let edges are = 3x, 4x and 5x

Diagonal of new cube formed = 12√3 cm (given)

Volume of new cube = Volume of effigy obtained after combining three cubes = (3x)^three + (4x) ^three + (5x) ³

= 216 x³ …(1)

New diagonal of a cube = √3a = 12√iii

or a = 12

Then, side of new cube is 12 cm.

Volume of cube with side 12 cm = (12)^3 …(2)

From equation (i) and (2), nosotros have

(12)^iii = 216 x³

Therefore, measure out of edges are :

3x = 3×2 = 6 cm

4x = 4×2 = 8cm

5x = five×2 = 10 cm

Question 3: If the perimeter of each face of a cube is 32 cm, find its lateral surface area. Notation 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 (given)

Let 'a' exist the edge of a cube.

We know, Perimeter of each face of a cube = 4a

⇒ 4a = 32

or a = viii

Side of a cube is 8 cm.

Now,

Lateral surface area of cube = 4a² = 4×viii² = 256 cm².

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RD Sharma Solutions for Form 9 Maths Chapter 18 Surface Area and Book of Cuboid and Cube

In the 18h Chapter of Grade 9 RD Sharma Solutions students will report important concepts listed below:

• Volume of a Cuboid
• Volume of a Cube

Benefits of using RD Sharma Solutions in exam preparation

• Students who exercise RD Sharma Solutions heighten the skills and confidence which are essential to secure high marks in examinations.
• Regular exercise of these solutions helps students to solve any type of problem in a comprehensive manner.
• Students who face difficulty in solving the bug are suggested to refer RD Sharma Solutions and get their doubts cleared.
• The solutions are well structured by professional teachers to meliorate conceptual knowledge among students.

Frequently Asked Questions on RD Sharma Solutions for Class ix Maths Chapter 18

How RD Sharma Solutions for Class 9 Maths Chapter 18 helpful for board test grooming?

RD Sharma Solutions for Form ix Maths Affiliate 18 is one of the best study sources that provides complete cognition of each and every concept for students. Practise is an essential job to larn and score well in Mathematics. Hence, an ample number of questions along with their solutions, shortcut techniques and detailed explanations are provided for practising any concept.

Which is the best source for Class 9 Maths lath examination training?

RD Sharma Textbooks are considered the all-time resource for CBSE board exam preparation. Students are required to go through RD Sharma Solutions for Class ix Maths Chapter 18 thoroughly before the concluding exams to score well and intensify their problem-solving abilities. After obtaining a concur on the concepts, students can move further with the sample papers and previous yr question papers to get amend acquainted with the test pattern.

What are the main concepts covered in RD Sharma Solutions for Class ix Maths Affiliate 18?

The main concepts covered in RD Sharma Solutions for Class 9 Maths Chapter xviii are as follows:
1. Volume of a Cuboid
2. Book of a Cube

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