Surface Area and Volume Class 9 With Answers PDF

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Surface Area and Volume Class 9 With Answers PDF Summary

Dear readers, today we are going to present Surface Area and Volume Class 9 PDF With Answers for all of you. Do you know about Surface Area and Volume? If not, then this article will prove to be very useful for you. Through this article, you can easily learn about Surface Area and Volume. It can be very helpful in your upcoming exam.
The amount of outer space that covers a three-dimensional shape is called it’s surface area. The total surface area of ​​a composite solid is the total surface area of the individual solids that make up the composite solid, except for the overlapping parts from each shape. Whereas the volume of a mixed solid is the sum of the volumes of the individual solids that make up the combined solid.
The surface area of ​​a solid shape is classified into three types of areas, namely lateral surface area, curved surface area, and total surface area. You can know about these three areas below on this page. The given information of Maths subject can be very beneficial in preparing and also getting good marks in the upcoming exam for the students.

Surface Area and Volume Class 9 PDF With Answers

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is open at the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box.

Ans:

We are given the length (l) of the box = 1.5 m.

The breadth (b) of box = 1.25 m.

The depth (h) of box = 0.65 m.

We are also told that the box is open at the top. So, we can write the area of the sheet required

So, the area of the sheet required

= 2lh+2bh+lb

[(× 1.5 × 0.65)+(× 1.25 × 0.65)+(1.5 × 1.25)]

= 1.95+1.625+1.875

= 5.45

The area of the sheet required is 5.45 m2.

(ii) The cost of the sheet for it, if a sheet measuring 1 m2 cost Rs. 20.

Ans: It is given that the cost of a sheet per m2 area = Rs. 20

Cost of sheet of 5.45 m2 area = Rs(20 × 5.45) = Rs. 109

So, the cost of the sheet required is Rs. 109.

2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2.

Ans:

We are given the following:

The length (l) of the room = 5 m.

The breadth (b) of the room = 4 m.

The height (h) of the room = 3 m.

We have to whitewash the walls of the room and also the ceiling.

The area to be whitewashed

= Area of walls + Area of ceiling

= 2lh+2bh+lb

[× 5 × 3+2 × 4 × 3+5 × 4] m2

(30+24+20) m2

= 74 m2

It is given that the cost of whitewashing per m2 area = Rs. 7.50

So, the cost of whitewashing 74 m2 area

= Rs(7.5 × 74)

= Rs. 555

The cost of whitewashing the walls of the given room is Rs. 555.

3. The floor of a rectangle hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of the hall.

(Hint: Area of four walls = Lateral surface area)

Ans:

Let us assume the following:

The length of the hall = l m

The breadth of the hall = b m

The height of the hall = h m

So, the area of the four walls will be

= 2lh+2bh

= 2(l+b)h

We are given the perimeter of the floor of the hall = 250 m. So, we can write the perimeter of the floor = 2(l+b) = 250 m.

Now we can rewrite the area of the four walls as the following.

Area of four walls = 2(l+b)h = 250h m2

The cost of painting per m2 area = Rs. 10

So, the cost of painting 250h m2 area = Rs(250h × 10) = Rs. 2500h

But in the question, it is given that the cost of painting the area of four walls is Rs. 15000

 2500h = 15000

h=6

Therefore, the height of the rectangular hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm ×10 cm ×7.5 cm can be painted out of this container?

Ans:

We will find the total surface area of one brick.

The total surface area of one brick

= 2(lb + bh + hl)

[2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)] cm2

= 2(225 + 75 + 168.75) cm2

(× 468.75) cm2

= 937.5 cm2

We are given that the paint in the container can paint out an area = 9.375 m2 = 93750 m2

Let us assume that n bricks can be painted out of the container.

So, the total area of n bricks (× 937.5) cm2 = 937.5n cm2

We can now equate the following:

93750 = 937.5n

n = 100

Therefore, we can paint out 100 bricks with the paint of the container.

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

i. Which box has the greatest lateral surface area and by how much?

Ans:

We are given the following:

Edge of the cubical box = 10 cm

Length (l) of the cuboidal box = 12.5 cm

Breadth (b) of the cuboidal box = 10 cm

Height (h) of the cuboidal box = 8 cm

Now we will calculate the lateral surface area of both the boxes.

The lateral surface area of the cubical box = 4(edge)2

= 4(10 cm)2

= 400 cm2

Lateral surface area of cuboidal box = 2[lh + bh]

[2(12.5 × 8 + 10 × 8)] cm2

(× 180) cm2

= 360 cm2

We can see that the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

The difference between the lateral surface areas = 400 cm2 – 360 cm2

= 40 cm2

Hence, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm2.

ii. Which box has the smaller total surface area and by how much?

Ans:

We will calculate the total surface area of both boxes.

Total surface area of cubical box = 6(edge)2

= 6(10 cm)2

= 600 cm2

Total surface area of cuboidal box = 2[lh + bh + lb]

= 2[12.5 × 8 + 10 × 8 + 12.5 × 10] cm2

= 2[100 + 80 + 125] cm2

= 610 cm2

We can see that the total surface area of the cubical box is smaller than the total surface area of the cuboidal box.

The difference between the total surface areas = 610 cm2 – 600 cm2

= 10 cm2

Hence, the total surface area of the cubical box is smaller than the total surface area of the cuboidal box by 10 cm2.

Surface Area and Volume Class 9 PDF With Answers: MCQs

Students of 9th Standard can choose the correct option and solve the MCQs on Surface areas and volumes.
1) The formula to find the surface area of a cuboid of length (l), breadth (b) and height (h) is:
a. lb+bh+hl
b. 2(lb+bh+hl)
c. 2(lbh)
d. lbh/2
Answer: b
2) The surface area of a cube whose edge equals 3cm is:
a. 62 sq.cm
b. 30 sq.cm
c. 54 sq.cm
d. 90 sq.cm
Answer: c
Explanation: Given, a = 3 cm
The surface area of the cube = 6a2
SA = 6 x 3 x 3 = 54 sq.cm
3) The volume of a hemisphere whose radius is r is:
a. 4/3 πr3
b. 4πr3
c. 2πr3
d. ⅔ π r3
Answer: d
4) The surface area of cuboid-shaped box having length=80 cm, breadth=40cm and height=20cm is:
a. 11200 sq.cm
b. 13000 sq.cm
c. 13400 sq.cm
d. 12000 sq.cm
Answer: a
Explanation: surface area of the box = 2(lb + bh + hl)
S.A. = 2[(80 × 40) + (40 × 20) + (20 × 80)]
= 2[3200 + 800 + 1600]
= 2 × 5600 = 11200 sq.cm.
5) The curved surface area of a right circular cylinder of the height of 14 cm is 88 cm2. The diameter of the base is:
a. 2 cm
b. 3cm
c. 4cm
d. 6cm
Answer: a
Explanation: Curved surface area of cylinder = 88 sq.cm
Height = 14 cm
2πrh = 88
r = 88/2πh
r=1 cm
Diameter = 2r = 2cm
6) The Curved surface area of a right circular cylinder is 4.4 sq. cm. The radius of the base is 0.7 cm. The height of the cylinder will be:
a. 2 cm
b. 3 cm
c. 1 cm
d. 1.5 cm
Answer: c
Explanation: Curved surface area of cylinder = 2πrh
2πrh = 4.4
h = 4.4/(2π x 0.7)
h = 1 cm
7) The diameter of the base of a cone is 10.5 cm, and its slant height is 10 cm. The curved surface area is:
a. 150 sq.cm
b. 165 sq.cm
c. 177 sq.cm
d. 180 sq.cm
Answer: b
Explanation: Diameter = 10.5, Radius = 10.5/2
Slant height, l = 10cm
Curved surface area of cone = πrl = π(5.25)(10)
CSA = 165 sq.cm
8) If the radius of a cylinder is 4cm and the height is 10cm, then the total surface area of a cylinder is:
a. 440 sq.cm
b. 352 sq.cm.
c. 400 sq. cm
d. 412 sq.cm
Answer: b
Explanation: Total Surface Area of a Cylinder = 2πr(r + h)
TSA = 2 x 22/7 x 4(4 + 10)
= (2x22x4x14)/7
= (2x22x4x2)
= 352 sq.cm
9) If the slant height of the cone is 21cm and the diameter of the base is 24 cm. The total surface area of a cone is:
a. 1200.77 sq.cm
b. 1177 sq.cm
c. 1222.77 sq.cm
d. 1244.57 sq.cm
Answer: d
Explanation: Total surface area = πr(l + r)
r = 24/2 = 12 cm
l = 21 cm
TSA = π(12)(21 + 12) = 1244.57 sq.cm
10) The surface area of a sphere of radius 14 cm is:
a. 1386 sq.cm
b. 1400 sq.cm
c. 2464 sq.cm
d. 2000 sq.cm
Answer: c
Explanation: Radius of sphere, r = 14 cm
Surface area = 4πr2
= 4 x 22/7 x (14)2 = 2464 sq.cm.
11) The radius of a sphere is 2r, then its volume will be
(a) (4/3) πr3
(b) 4πr3
(c) (8/3) πr3
(d) (32/3) πr3
Answer: d
Explanation:
Given : r=2r
The volume of sphere = (4/3)πr= (4/3)π(2r)3
V = (4/3)π(8r3) = (32/3)πr3.
12) The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is pumped into it. The ratio of the surface areas of the balloon in the two cases is
(a) 1:4
(b) 1:3
(c) 2:3
(d) 2:1
Answer: a
Explanation:
We know that the total surface area of the hemisphere = 3πr2 square units.
If r= 6cm, then TSA = 3π(6)2 = 108π
If r = 12 cm, then TSA = 3π(12)2= 432π
Then the ratio = (108π)/(432π)
Ratio = ¼, which is equal to 1:4.
13) The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5m) is
(a) 15m
(b) 16m
(c) 10m
(d) 12m
Answer: a
Explanation:
Given: l=10m, b= 10m, h= 5m
The length of the longest pole = √[102+102+52]
= √(100+100+25) = √225 = 15 m.
14) A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2. 4 cm
(d) 1.6 cm
Answer: b
Explanation:
Given that the height of the cone = 8.4 cm
Radius of cone = 2.1 cm
Also, given that the volume of cone = volume of a sphere
(⅓)πr2h = (4/3)πr3
(⅓)π(2.1)2(8.4) = (4/3)πr3
37.044= 4r3
r3= 37.044/4
r3= 9.261
r = 2.1
Therefore, the radius of the sphere is 2.1 cm.
15) The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit that is 16 m long, 12m wide and 4 m deep is
(a) 1900
(b) 1920
(c) 1800
(d) 1840
Answer: b
Explanation:
Volume of Plank = 400 cm×50cm×20cm=400000cm3
Volume of pits = 1600cm×1200cm×400cm = 768000000cm3
Number of planks = Volume of planks/Volume of pits
= 768000000/400000
Hence, the number of pits = 1920
16) The total surface area of a cube is 96 cm2. The volume of the cube is:
(a) 8 cm3
(b) 512 cm3
(c) 64 cm3
(d) 27 cm3
Answer: c
Explanation:
We know that the TSA of the cone = 6a2.
6a2 = 96 cm2
a2 = 96/6 = 16
a =4 cm
The volume of cone = a3 cubic units
V = 43 = 64cm3.
17) In a cylinder, the radius is doubled and height is halved, the curved surface area will be
(a) Halved
(b) Doubled
(c) Same
(d) Four times
Answer: c
Explanation:
We know that the curved surface area of a cylinder is 2πrh
Given that, r = 2R, h= H/2
Hence, the CSA of new cylinder = 2π(2R)(H/2) = 2πRH
Therefore, the answer is “Same”.
18) The lateral surface area of a cube is 256 m2. The volume of the cube is
(a) 512 m3
(b) 64 m3
(c) 216 m3
(d) 256 m3
Answer: a
Explanation:
The lateral surface area of the cube = 4a2
4a2= 256
a2 = 256/4 =64
a = 8 m
Hence, the volume of cube = a3 cube units
V = 8 = 512 m3.
19) The total surface area of a cone whose radius is r/2 and slant height 2l is
(a) 2πr(l+r)
(b) πr(l+(r/4))
(c) πr(l+r)
(d) 2πrl
Answer: b
Explanation:
The total surface area of the cone = πr(l+r) square units.
If r = r/2 and l= 2l, then the TSA of the cone becomes,
TSA of cone = π(r/2)[(2l+(2/r)]
=π[(RL)+(r2/4)]
TSA of new cone =πr[l+(r/4)]
20) The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3. The ratio of their volumes is:
(a) 10: 17
(b) 20: 27
(c) 17: 27
(d) 20: 37
Answer: b
Explanation:
Given that, the radii of two cylinders are in the ratio of 2:3
Hence, r1= 2r, r2 = 3r
Also, given that, the height of two cylinders is in the ratio of 5:3.
Hence, h1 = 5h, h2 = 3h
The ratio of the volume of two cylinders = V1/V2
= πr12h1/πr22h2
= [(2r)2(5h)]/[(3r)2(3h)]
Ratio of their volumes =(20r2h)/(27r2h) = 20/27 = 20: 27.
Stay tuned with BYJU’S – The Learning App and download the app today to get more class-wise concepts.

Three Different Areas

Lateral Surface Area: The lateral surface area is known as the area of all the faces except the bottom and top faces or bases.
Curved Surface Area: The curved surface area is known as the area of all the curved regions of the solid.
Total Surface Area: The total surface area is the area of all the faces (including top and bottom faces) of the solid object.

Surface Area and Volume

Some of the major shapes to which the students of class 9 are introduced are cuboid, cylinder, cone and sphere. We will discuss each one of them one by one.
Cuboid: The shape of a cuboid has rectangular faces which are six in number and are placed right-angled to each other. The formula of the total surface area of a cuboid is equal to the total area of all the six rectangles.
Cube: A cube is similar to the cuboid which has all six surfaces, that is the length, breadth and height are equal. The number of edges and vertices are 12 and 8 respectively. The total surface area of a cube is 6s2. This is because here length= breadth= height (let us assume it s cm)
Total surface area = 2*3 s2
Right Circular Cylinder: Right circular cylinder is a surface in which two curved surfaces are present to connect the two parallel circular bases. The two bases will be exactly over one another. The formula of the total surface area of a right circular cylinder is as follows:
TSA = 2π × r × h + 2 × πr2
⇒ TSA = 2πr(h + r)
Here r is the radius of the surface and h is the height of the cylinder. If we have to find the curved surface area then it will be 2 RH
Right Circular Cone: If we first take out the relation between the slant height (l) and height of the cone then by using the Pythagoras theorem it will be l2=h2+r2. The curved surface area of the right circular cone will be the total sum of the small triangles formed. The final formula for the CSA will be rl
The total surface area of the right circular cone will be the total of CSA and the area of the base. As mentioned above the CSA will be rl and the area of the base will be r²
Therefore the TSA of the right circular cone will be r(l+r).
Sphere: A three-dimensional closed solid figure is a sphere. In a sphere the distance of the surface from a common point which is the centre is equal. In a sphere, the total surface area and the curved surface area are equal which is 4r2

The volume of the Surfaces

Now that we have taken a look at the total surface area and the curved or lateral surface area of the three-dimensional structure.
Now let us look at the volume of these surfaces:

  • The formula for the cuboid is lbh.
  • The formula for the volume of a cube is the base area multiplied by the height. As all the surfaces of the cube are equal. The formula generated is a³
  • The formula for the volume of a right circular cylinder is the base area multiplied by the height. The formula formed will be r2h
  • The formula for the volume of a right circular cone is ar2h.

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