NCERT Class 9 Maths: Surface Area and Volume Ex 13.4

Welcome, Class 9 students, to an exciting journey into the world of three-dimensional shapes with YoLearn.ai! In this deep dive into NCERT Exercise 13.4 from the chapter "Surface Area and Volume," we will explore the fascinating properties of spheres and hemispheres. These shapes are all around us, from the balls we play with to the domes of famous buildings. Understanding their surface area is crucial not just for your exams, but also for appreciating the geometry that governs our physical world. By the end of this session, you will not only be able to confidently calculate the surface area of spheres and hemispheres but also solve complex problems involving these shapes, mastering the concepts required for your CBSE Class 9 Maths examination. Get ready to strengthen your conceptual understanding and ace your board exams!

Understanding Spheres and Hemispheres: Definitions and Formulas

A sphere is a perfectly round three-dimensional object, where every point on its surface is equidistant from its center. Think of a perfectly round cricket ball or a globe. Its surface is continuous and has no flat faces or edges. To calculate the amount of material needed to cover its entire outer surface, we use the concept of Surface Area. For a sphere with radius 'r', its surface area (SA) is given by the formula:

Surface Area of a Sphere = 4πr²

A hemisphere is exactly half of a sphere. Imagine cutting a sphere exactly through its center. The resulting shape will have a curved surface and a flat circular base. This distinction is very important when calculating its surface area.

There are two types of surface areas for a hemisphere:

  1. Curved Surface Area (CSA) of a Hemisphere: This refers only to the curved portion, like the outer shell of a bowl. For a hemisphere with radius 'r', its Curved Surface Area is half the surface area of a full sphere:

CSA of a Hemisphere = 2πr²

  1. Total Surface Area (TSA) of a Hemisphere: This includes both the curved surface and the flat circular base. The area of the circular base is πr². Therefore, the Total Surface Area is the sum of the Curved Surface Area and the area of the base:

TSA of a Hemisphere = Curved Surface Area + Area of Circular Base = 2πr² + πr² = 3πr²

Understanding these formulas and when to apply each one is key to mastering Exercise 13.4. Always pay close attention to whether the problem asks for the surface area of a sphere, or the curved or total surface area of a hemisphere.

Worked Examples: Applying Surface Area Formulas

  • Example 1: Find the surface area of a sphere of radius 10.5 cm. Step 1: Identify the given information. Radius (r) = 10.5 cm. We need to find the Surface Area of a Sphere. Step 2: Recall the formula for the Surface Area of a Sphere: SA = 4πr². Step 3: Substitute the value of r and π (use π = 22/7 for calculations unless otherwise specified). SA = 4 × (22/7) × (10.5)² SA = 4 × (22/7) × (10.5 × 10.5) SA = 4 × (22/7) × 110.25 SA = 88/7 × 110.25 SA = 88 × 15.75 (since 110.25 / 7 = 15.75) SA = 1386 cm² Final Answer: The surface area of the sphere is 1386 cm².
  • Example 2: A hemispherical bowl is made of brass. The inner diameter of the bowl is 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm². Step 1: Identify given information. Inner diameter = 10.5 cm. Rate of tin-plating = ₹16 per 100 cm². Step 2: Calculate the inner radius (r). Radius = Diameter / 2 = 10.5 / 2 = 5.25 cm. Step 3: Since tin-plating is done on the inside of the bowl, we need to find the Curved Surface Area (CSA) of the hemisphere. CSA of hemisphere = 2πr². Step 4: Substitute the values of r and π (use π = 22/7). CSA = 2 × (22/7) × (5.25)² CSA = 2 × (22/7) × (5.25 × 5.25) CSA = 44/7 × 27.5625 CSA = 44 × 3.9375 (since 27.5625 / 7 = 3.9375) CSA = 173.25 cm² Step 5: Calculate the cost of tin-plating. Cost for 100 cm² = ₹16 Cost for 1 cm² = ₹16 / 100 = ₹0.16 Total cost = CSA × Cost per cm² = 173.25 × 0.16 Total cost = ₹27.72 Final Answer: The cost of tin-plating the inside of the bowl is ₹27.72.
  • Example 3: Find the Total Surface Area of a hemisphere of radius 7 cm. Step 1: Identify the given information. Radius (r) = 7 cm. We need to find the Total Surface Area of a hemisphere. Step 2: Recall the formula for the Total Surface Area of a Hemisphere: TSA = 3πr². Step 3: Substitute the value of r and π (use π = 22/7). TSA = 3 × (22/7) × (7)² TSA = 3 × (22/7) × (7 × 7) TSA = 3 × 22 × 7 (one 7 cancels out with the denominator 7) TSA = 66 × 7 TSA = 462 cm² Final Answer: The total surface area of the hemisphere is 462 cm².

Exam Tip: Avoiding Common Mistakes in Surface Area Calculations

To score full marks in questions involving surface area of spheres and hemispheres, pay close attention to these common pitfalls:

  1. Diameter vs. Radius: Always double-check if the problem provides the diameter or the radius. If diameter (d) is given, remember to convert it to radius (r) using the relation r = d/2 before applying any formula. This is one of the most frequent mistakes students make.
  2. CSA vs. TSA for Hemispheres: For hemispheres, carefully read whether the question asks for the Curved Surface Area (CSA) or the Total Surface Area (TSA). CSA only includes the curved part (2πr²), while TSA includes the curved part and the flat circular base (3πr²).
  3. Units: Ensure that you always write the correct units for your final answer. Since it's surface area, the units will always be square units (e.g., cm², m²). Incorrect or missing units can lead to deduction of marks.
  4. Value of π: Use the value of π specified in the question (e.g., 22/7 or 3.14). If not specified, 22/7 is generally preferred for calculations involving multiples of 7 for radius or diameter.

Practice Questions with Solutions

  • Q: Find the surface area of a sphere of diameter 14 cm. A: Step 1: Identify the given information. Diameter (d) = 14 cm. Step 2: Calculate the radius (r). r = d/2 = 14/2 = 7 cm. Step 3: Apply the formula for the Surface Area of a Sphere: SA = 4πr². SA = 4 × (22/7) × (7)² SA = 4 × (22/7) × 49 SA = 4 × 22 × 7 SA = 88 × 7 = 616 cm². Final answer: The surface area of the sphere is 616 cm².
  • Q: A spherical balloon's radius increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of surface areas of the balloon in the two cases. A: Step 1: Identify the initial radius (r1) = 7 cm and the final radius (r2) = 14 cm. Step 2: Calculate the initial surface area (SA1) of the balloon. SA1 = 4πr1² = 4π(7)² = 4π(49) = 196π cm². Step 3: Calculate the final surface area (SA2) of the balloon. SA2 = 4πr2² = 4π(14)² = 4π(196) = 784π cm². Step 4: Find the ratio of SA1 to SA2. Ratio = SA1 / SA2 = (196π) / (784π) Ratio = 196 / 784 = 1 / 4. Final answer: The ratio of the surface areas is 1:4.
  • Q: A hemispherical dome needs to be painted. If the radius of the base of the dome is 21 metres, find the cost of painting it at the rate of ₹5 per square metre. A: Step 1: Identify the given information. Radius (r) = 21 m. Cost rate = ₹5 per m². Step 2: Since it's a dome, only the curved surface needs to be painted. Calculate the Curved Surface Area (CSA) of the hemisphere. CSA = 2πr² = 2 × (22/7) × (21)² CSA = 2 × (22/7) × (21 × 21) CSA = 2 × 22 × 3 × 21 (as 21/7 = 3) CSA = 44 × 63 = 2772 m². Step 3: Calculate the total cost of painting. Cost = CSA × Rate = 2772 × 5 = ₹13860. Final answer: The cost of painting the dome is ₹13860.
  • Q: The radius of a sphere is 3.5 cm. Find its surface area. (Use π = 22/7) A: Step 1: Given radius (r) = 3.5 cm. Step 2: Formula for surface area of a sphere: SA = 4πr². Step 3: Substitute values. SA = 4 × (22/7) × (3.5)² SA = 4 × (22/7) × (3.5 × 3.5) SA = 4 × (22/7) × 12.25 SA = 4 × 22 × 1.75 (since 12.25 / 7 = 1.75) SA = 88 × 1.75 SA = 154 cm². Final answer: The surface area of the sphere is 154 cm².
  • Q: A hemispherical tank has an inner radius of 1.4 m. It is to be covered from the outside. Find the total area to be covered. (Use π = 22/7) A: Step 1: Given inner radius (r) = 1.4 m. Step 2: Since the tank is to be covered from the outside, we need to find the Total Surface Area (TSA) of the hemisphere (assuming it includes the base if covered). TSA = 3πr². Step 3: Substitute values. TSA = 3 × (22/7) × (1.4)² TSA = 3 × (22/7) × (1.4 × 1.4) TSA = 3 × (22/7) × 1.96 TSA = 3 × 22 × 0.28 (since 1.96 / 7 = 0.28) TSA = 66 × 0.28 TSA = 18.48 m². Final answer: The total area to be covered is 18.48 m².

Frequently Asked Questions

What is the difference between the Curved Surface Area (CSA) and Total Surface Area (TSA) of a hemisphere?

The Curved Surface Area (CSA) of a hemisphere refers only to its rounded, outer surface (2πr²). The Total Surface Area (TSA) includes both this curved part and the flat circular base, making it 3πr².

When should I use 22/7 for π and when should I use 3.14?

You should use the value of π specified in the question. If no value is specified, 22/7 is generally a good choice, especially if the radius or diameter is a multiple of 7, as it simplifies calculations. Use 3.14 if the problem specifies or if dealing with values that don't easily divide by 7.

Why is the surface area of a sphere 4πr²?

While a formal derivation involves calculus, conceptually, one can imagine covering a sphere with four circles of the same radius 'r' as the sphere. Ancient mathematicians like Archimedes proved that the surface area of a sphere is indeed equal to the area of four great circles, each with area πr².

Can I use the formulas for real-world objects like a football or a dome?

Yes, these formulas are directly applicable to many real-world objects that approximate spheres or hemispheres. For example, a football can be considered a sphere to calculate its surface area, and the surface of a dome can be calculated using the hemispherical surface area formulas.