Surface Area & Volume: Solved Examples for Class 9-10

Suno, do you ever feel a shiver down your spine when you see a question about finding the surface area of a combined solid? Or maybe you just stare blankly at problems involving converting a sphere into smaller cones? You're not alone, yaar!

Surface Area and Volume (Mensuration, as we call it in Class 9-10) can feel tricky. It's not just about memorizing formulas; it's about understanding when and how to apply them. But trust me, with the right approach and enough practice, you can totally ace this chapter!

This article is your personal guide to cracking Mensuration. We'll go through some common, yet challenging, solved examples that cover cylinders, cones, spheres, hemispheres, and even combinations of solids. We'll also dive into the conversion of solids, which is a favourite topic for board examiners. So, ready to turn those frowns into confident smiles? Let's get started!

Accha, ever wondered why we even learn all this? It's not just for exams, you know! Mensuration is everywhere around us.

Think about architects designing buildings, they need to calculate the volume of concrete and the surface area of walls to estimate paint. Engineers use it to design pipelines and storage tanks. Even your local chaiwala uses it when deciding the size of his kulhad! From packaging design to space exploration, these concepts are fundamental. It’s a foundational skill for many future careers, especially if you're eyeing fields like engineering, design, or even data science, where spatial reasoning is key.

Let's kick things off with a classic combination problem. These are super common in both CBSE (NCERT Chapter 13) and ICSE exams.

Problem: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1 \text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

Solution:

Step 1: Identify the components and given values.
The solid is a combination of a cone and a hemisphere.
Radius of hemisphere, $r = 1 \text{ cm}$.
Radius of cone, $r = 1 \text{ cm}$ (since both radii are equal).
Height of cone, $h = r = 1 \text{ cm}$.

Step 2: Write down the formulas for the volume of each component.
Volume of a cone, $V_{\text{cone}} = \frac{1}{3} \pi r^2 h$
Volume of a hemisphere, $V_{\text{hemisphere}} = \frac{2}{3} \pi r^3$

Step 3: Substitute the given values into the formulas.
$V_{\text{cone}} = \frac{1}{3} \pi (1)^2 (1) = \frac{1}{3} \pi \text{ cm}^3$
$V_{\text{hemisphere}} = \frac{2}{3} \pi (1)^3 = \frac{2}{3} \pi \text{ cm}^3$

Step 4: Add the volumes to find the total volume of the solid.
Total Volume, $V_{\text{solid}} = V_{\text{cone}} + V_{\text{hemisphere}}$
$V_{\text{solid}} = \frac{1}{3} \pi + \frac{2}{3} \pi$
$V_{\text{solid}} = \left(\frac{1}{3} + \frac{2}{3}\right) \pi$
$V_{\text{solid}} = \frac{3}{3} \pi$
$V_{\text{solid}} = \pi \text{ cm}^3$

So, the volume of the solid is $\pi \text{ cm}^3$. This type of question often appears in the 3-mark section of CBSE board papers.

Surface area problems, especially involving paint or covering, are another favourite. Let's look at one involving a hemisphere.

Problem: A hemispherical bowl of internal radius $9 \text{ cm}$ is full of liquid. This liquid is to be filled into cylindrical bottles of radius $1.5 \text{ cm}$ and height $4 \text{ cm}$. How many bottles are needed to empty the bowl?

Solution:

Step 1: Determine the volume of liquid in the hemispherical bowl.
Internal radius of hemispherical bowl, $R = 9 \text{ cm}$.
Volume of hemisphere, $V_{\text{hemisphere}} = \frac{2}{3} \pi R^3$
$V_{\text{hemisphere}} = \frac{2}{3} \pi (9)^3 = \frac{2}{3} \pi (729) = 2 \pi (243) = 486 \pi \text{ cm}^3$

Step 2: Determine the volume of one cylindrical bottle.
Radius of cylindrical bottle, $r = 1.5 \text{ cm}$.
Height of cylindrical bottle, $h = 4 \text{ cm}$.
Volume of a cylinder, $V_{\text{cylinder}} = \pi r^2 h$
$V_{\text{cylinder}} = \pi (1.5)^2 (4) = \pi (2.25) (4) = 9 \pi \text{ cm}^3$

Step 3: Calculate the number of bottles needed.
Number of bottles = $\frac{\text{Volume of hemispherical bowl}}{\text{Volume of one cylindrical bottle}}$
Number of bottles = $\frac{486 \pi}{9 \pi}$
Number of bottles = $\frac{486}{9} = 54$

Thus, $54$ cylindrical bottles are needed to empty the hemispherical bowl. This is a common conversion of solids question, often carrying 4 marks in your board exams.

Melting and recasting problems are a bit different because the volume remains constant. This is a key concept for both CBSE and ICSE students. Let's tackle one.

Problem: A metallic sphere of radius $4.2 \text{ cm}$ is melted and recast into the shape of a cylinder of radius $6 \text{ cm}$. Find the height of the cylinder.

Solution:

Step 1: Understand the principle.
When a solid is melted and recast into another solid, its volume remains unchanged.
So, Volume of sphere = Volume of cylinder.

Step 2: Write down the formulas and given values.
Radius of sphere, $R = 4.2 \text{ cm}$.
Volume of sphere, $V_{\text{sphere}} = \frac{4}{3} \pi R^3$
Radius of cylinder, $r = 6 \text{ cm}$.
Height of cylinder, $h = ?$
Volume of cylinder, $V_{\text{cylinder}} = \pi r^2 h$

Step 3: Equate the volumes and solve for the unknown.
$\frac{4}{3} \pi R^3 = \pi r^2 h$
Cancel $\pi$ from both sides:
$\frac{4}{3} R^3 = r^2 h$

Substitute the given values:
$\frac{4}{3} (4.2)^3 = (6)^2 h$
$\frac{4}{3} (4.2 \times 4.2 \times 4.2) = 36 h$

Let's simplify:
$4 \times (1.4 \times 4.2 \times 4.2) = 36 h$
$4 \times 24.696 = 36 h$
$98.784 = 36 h$
$h = \frac{98.784}{36}$
$h = 2.744 \text{ cm}$

Therefore, the height of the cylinder is $2.744 \text{ cm}$. These problems are often worth 3-4 marks and require careful calculation, so don't rush!

Alright, let's quickly recap what we've learned to conquer Surface Area and Volume:

* Formulas are your friends: Know them inside out, but also understand their derivation.
* Visualize: Always draw diagrams for complex shapes or combined solids.
* Break it Down: For combination problems, find the surface area or volume of individual components first.
* Volume Conservation: Remember that volume remains constant during melting and recasting.
* Consistent Practice: Regular problem-solving is the only way to build speed and accuracy.
* Board-Specific Focus: Tailor your preparation to CBSE or ICSE patterns, but ensure strong fundamentals for both.

With these tips and the practice you'll put in, you'll be solving Mensuration problems like a pro in no time! All the best for your exams!