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Important Questions for Class 9 Maths Chapter 13 Surface Areas Volumes

Here, we have provided Important Questions for Class 9 Maths Chapter 13. Students can view these Important Questions for Class 9 Maths Chapter 13 before exams for better understanding of the subject.
authorImageAnanya Gupta24 Sept, 2025
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Important Questions for Class 9 Maths Chapter 13

Important Questions for Class 9 Maths Chapter 13: In Chapter 13 of Class 9 Maths students learn about surface areas and volumes of different shapes like cubes, cuboids, cylinders, cones and spheres. This chapter helps students understand how to use formulas to find out how much space these shapes take up and how much area their surfaces cover.

Important questions usually involve calculating the surface area and volume of these shapes and applying these concepts to real-life situations, like finding out how much paint is needed to cover a wall or how much water a container can hold. Practicing these questions helps students improve their problem-solving skills and get ready for their exams

 Class 9 Maths Chapter 13 Important Questions

The important questions for Class 9 Maths Chapter 13 are created by by folllowing latest CBSE Class 9 Maths Syllabus focus on key concepts related to surface areas and volumes. These questions are created to enhance students understanding of how to calculate the surface area and volume of various three-dimensional shapes, such as cubes, cuboids, cylinders, cones and spheres.

By tackling these questions students can solidify their grasp of formulas and improve their problem-solving abilities. The expert-created questions not only prepare students for their exams but also help them apply these mathematical concepts to real-world scenarios making their learning experience more relevant and engaging.

Important Questions for Class 9 Maths Chapter 13 PDF

The PDF link for the Important Questions for Class 9 Maths Chapter 13 is available below.  Students can use this PDF to practice and reinforce their understanding of the concepts discussed in the chapter. By working through these important questions, learners can prepare effectively for their exams and enhance their problem-solving skills. Download the PDF now to access valuable practice material that will help in mastering this crucial chapter in mathematics.

Important Questions for Class 9 Maths Chapter 13 PDF

 Surface Areas Volumes Important Questions Class 9 Maths

Here we have provided Important Questions CBSE Class 9 Maths Chapter 13 Surface Areas Volumes-
Q.1: Hameed has built a cubical water tank with a lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see in the figure below).

Find how much he would spend on the tiles if the cost of the tiles is Rs.360 per dozen.
Solution: Given, Edge of the cubical tank (a) = 1.5 m = 150 cm So, surface area of the tank = 5 × 150 × 150 cm 2 The measure of side of a square tile = 25 cm Area of each square tile = side × side = 25 × 25 cm 2 Required number of tiles = (Surface area of the tank)/(area of each tile) = (5 × 150 × 150)/(25 × 25) = 180 Also, given that the cost of the tiles is Rs. 360 per dozen. Thus, the cost of each tile = Rs. 360/12 = Rs. 30 Hence, the total cost of 180 tiles = 180 × Rs. 30 = Rs. 5400

Q.2: The paint in a certain container is sufficient to paint an area equal to 9.375 sq.m. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution: Given, Dimensions of the brick = 22.5 cm × 10 cm × 7.5 cm Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm Surface area of 1 brick  = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm 2 = 2(225 + 75 + 168.75) cm 2 = 2 x 468.75 cm 2 = 937.5 cm 2 Area that can be painted by the container = 9.375 m 2 (given) = 9.375 × 10000 cm 2 = 93750 cm 2 Thus, the required number of bricks = (Area that can be painted by the container)/(Surface area of 1 brick) = 93750/937.5 = 937500/9375 = 100

Q.3: 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 sq.m.
Solution: Given, Length of the room (l) = 5 m Breadth of the room (b) = 4 m Height of the room (h) = 3 m Area of walls of the room = Lateral surface area of cuboid = 2h(l + b) = 2 × 3(5 + 4) = 6 × 9 = 54 sq.m Area of ceiling = Area of base of the cuboid = lb = 5 × 4 = 20 sq.m Area to be white washed = (54 + 20) sq.m = 74 sq.m Given that, the cost of white washing 1 sq.m = Rs. 7.50 Therefore, the total cost of white washing the walls and ceiling of the room = 74 × Rs. 7.50 = Rs. 555

 Q.4: The curved surface area of a right circular cylinder of height 14 cm is 88 sq.cm. Find the diameter of the base of the cylinder.
Solution: Let d be the diameter and r be the radius of a right circular cylinder. Given, Height of cylinder (h) = 14 cm Curved surface area of right circular cylinder = 88 cm 2 ⇒ 2πrh = 88 cm 2 ⇒ πdh = 88 cm 2 (since d = 2r) ⇒ 22/7 x d x 14 cm = 88 cm 2 ⇒ d = 2 cm Hence, the diameter of the base of the cylinder is 2 cm.

Q.5: Curved surface area of a right circular cylinder is 4.4 sq.m. If the radius of the base of the cylinder is 0.7 m, find its height. Solution: Let h be the height of the cylinder. Given, Radius of the base of the cylinder (r) = 0.7 m Curved surface area of cylinder = 4.4 m 2 Thus, 2πrh = 4.4 2 × 3.14 × 0.7 × h = 4.4 4.4 × h = 4.4 h = 4.4/4.4 h = 1 Therefore, the height of the cylinder is 1 m.

Q.6: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution: Given, Length of the cylindrical pipe = h = 28 m Diameter of the pipe = 5 cm Now, the radius of piper (r) = 5/ 2 cm = 2.5 cm = 0.025 m Total radiating surface in the system = Total surface area of the cylinder = 2πr(h + r) = 2 × (22/7) × 0.025 (28 + 0.025) m 2 = (44 x 0.025 x 28.025)/7 m 2 = 4.4 m 2 (approx)

Q.7: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. (Take π = 3.14)
Solution: Given Height of a cone (h) = 16 cm Radius of the base (r) = 12 cm Now, Slant height of cone (l) = √(r 2 + h 2 ) = √(256 + 144) = √400 = 20 cm Curved surface area of cone = πrl = 3.14 × 12 × 20 cm 2 = 753.6 cm 2 Total surface area = πrl + πr 2 = (753.6 + 3.14 × 12 × 12) cm 2 = (753.6 + 452.16) cm 2 = 1205.76 cm 2

Q.8: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution: Given, Diameter of the cone = 24 m Radius of the cone (r) = 24/2 = 12 m Slant height of the cone (l) = 21 m Total surface area of a cone = πr(l + r) = (22/7) × 12 × (21 + 12) = (22/7) × 12 × 33 = 1244.57 m 2

Q.9: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs.210 per 100 sq.m.
Solution: Given, Slant height of a cone (l) = 25 m Diameter of the base of cone = 2r = 14 m ∴ Radius = r = 7 m Curved Surface Area = πrl = (22/7) x 7 x 25 = 22 × 25 = 550 sq.m Also, given that the cost of white-washing 100 sq.m = Rs. 210 Hence, the total cost of white-washing for 550 sq.m = (Rs. 210 × 550)/100 = Rs. 1155

Q.10: The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Solution: Given, Diameter of the sphere = 7 m Radius (r) = 7/2 = 3.5 m Now, the riding space available for the motorcyclist = Surface area of the sphere = 4πr 2 = 4 × (22/7) × 3.5 × 3.5 = 154 m 2

Q.11: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution: Given, Radius of balloon = r =  7 cm Radius of pumped balloon = R = 14 cm Ratio of surface area = (TSA of balloon with r = 7 cm)/(TSA of balloon with R = 14 cm) = (4πr 2 )/(4πR 2 ) = r 2 /R 2 = (7) 2 /(14) 2 = 49/196 = 1/4 Hence, the ratio of surface areas of the balloon in the two cases is 1 : 4.

Q.12: 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?
Solution: Given, Depth of the river (h) = 3 m Width of the river (w) = 40 m Flow rate of water = 2 km/hr i.e. Flow of water in 1 hour = 2 km = 2000 m Flow of water in 1 minute = 2000/60 = 100/3 m Thus, length (l) = 100/3 m Volume of water falling into the sea in 1 minute = Volume of cuboid with dimension l, w, h = l × w × h = (100/3) × 40 × 3 = 4000 m 3 = 4000 x 1000 L = 4000000 L

 Q.13: A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution: Given, Diameter of the pencil = 7  mm Radius of the pencil (R) = 7/2 mm Diameter of the graphite cylinder = 1 mm Radius of the graphite (r) = 1/2 mm Height (h) = 14 cm = 140 mm (since 1 cm = 10 mm) Volume of a cylinder = πr²h Volume of graphite cylinder = πr2h = (22/7) × (1/2) × (1/2) × 140 = 110 mm³ Volume of pencil = πR²h = (22/7) × (7/2) × (7/2) × 140 = 490 × 11 = 5390 mm² Volume of wood = Volume of pencil – Volume of graphite = 5390- 110 = 5280 mm³ = 5280/1000 (since 1 mm³ = 1/1000 cm³) = 5.28 cm³

 Q.14: Meera has a piece of canvas whose area is 551 m 2 . She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m 2 , find the volume of the tent that can be made with it.
Solution: Given, Area of the canvas = 551 m 2 Area of the canvas lost in wastage = 1 m 2 Thus, the area of canvas available for making the tent = (551 – 1) m 2 = 550 m 2 Now, the surface area of the tent = 550 m 2 The required base radius of the conical tent = 7 m Curved surface area of tent = 550 m 2 That means, πrl = 550 (22/7) × 7 × l = 550 l = 550/22 l = 25 m Now, l 2 = h 2 + r 2 h 2 = l 2 – r 2 = (25) 2 – (7) 2 = 625 – 49 = 576 h = 24 m So, the volume of the conical tent = (1/3)πr 2 h = (1/3) × (22/7) × 7 × 7 × 24 = 1232 m 3

 Q.15: A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm 3 ) is needed to fill this capsule?
Solution: Given, Diameter of capsule = 3.5 mm Radius of capsule = (r) = 3.5/2 = 1.75 mm Volume of spherical capsule = (4/3)πr 3 = (4/3) × (22/7) × 1.75 × 1.75 × 1.75 = 22.458 mm 3 Therefore, the volume of the capsule is 22.46 mm 3 approx.

 Q.16: Calculate the amount of ice-cream that can be put into a cone with base radius 3.5 cm and height 12 cm.
Solution: Given, Base radius = r = 3.5 cm Height = h = 12 cm The amount of ice-cream that can be put into a cone = Volume of cone = (1/3)πr 2 h = (1/3) × (22/7) × 3.5 × 3.5 × 12 = 154 cm3

Q.17: A spherical ball is divided into two equal halves. Given that the curved surface area of each half is 56.57 cm, what will be the volume of the spherical ball?
Solution: Given, Curved surface area of of half of the spherical ball = 56.57 cm 2 (1/2) 4πr 2 = 56.57 2 × 3.14 × r 2 = 56.57 r 2 = 56.57/6.28 r 2 = 9 (approx) r = 3 cm Now, Volume of spherical ball = (4/3)πr 3 = (4/3) × 3.14 × 3 × 3 × 3 = 113.04 cm 3

Benefits of Practicicng Important Questions for Class 9 Maths Chapter 13

Practicing important questions for Class 9 Maths Chapter 13 on Surface Areas and Volumes provide several key benefits for students:
Conceptual Clarity : Regular practice helps students understand the concepts of surface areas and volumes of various geometric shapes, including cubes, cuboids, cylinders, cones and spheres.
Enhanced Problem-Solving Skills : By working through different types of questions, students develop analytical and critical thinking skills, making them better equipped to tackle complex problems.
Boosted Confidence : Familiarity with important questions increases students confidence levels as they approach their exams, reducing anxiety and enhancing overall performance.
Time Management Skills : Practicing with a variety of questions allows students to improve their time management skills, helping them complete their exams within the given time frame.
Exam Preparedness : Focusing on important questions ensures that students are well-prepared for their exams, as these questions often mirror the types of problems they will face on the exam.

Important Questions for Class 9 Maths Chapter 13 FAQs

What is the difference between surface area and volume?

Surface area refers to the total area of the outer surface of a three-dimensional shape, while volume measures the amount of space occupied by the shape.

Can surface area and volume be the same for any shape?

Generally, surface area and volume are different measurements, but there may be specific instances where the numerical values coincide for particular shapes with specific dimensions.

Why is it important to learn about surface areas and volumes?

Understanding surface areas and volumes is essential in various fields, including engineering, architecture and everyday problem-solving related to capacity and material use.

How can I improve my skills in calculating surface areas and volumes?

Practice solving problems regularly, refer to geometry textbooks or online resources and work on sample questions to strengthen your understanding of these concepts.
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