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CBSE Class 10 Maths Chapter 12: Surface Areas and Volumes

📖 Formula Sheet ✏️ NCERT Solutions 📥 PDF Notes

1. Ultimate 3D Formula Sheet Table

Students look for this everywhere. Having this complete summary table on your page is awesome for search ranking:

Solid Figure Curved / Lateral Surface Area (CSA/LSA) Total Surface Area (TSA) Volume
Cuboid$2h(l + b)$$2(lb + bh + hl)$$l \times b \times h$
Cube$4a^2$$6a^2$$a^3$
Right Circular Cylinder$2\pi rh$$2\pi r(r + h)$$\pi r^2 h$
Right Circular Cone$\pi rl$
($l = \sqrt{r^2 + h^2}$)		$\pi r(l + r)$$\frac{1}{3}\pi r^2 h$
Sphere$4\pi r^2$$4\pi r^2$$\frac{4}{3}\pi r^3$
Hemisphere$2\pi r^2$$3\pi r^2$$\frac{2}{3}\pi r^3$

2. Combination of Solids Concept

In Class 10, questions are based on figures joined together (e.g., a cone placed on top of a hemisphere to make a toy):

⚠️ The Golden Scroll Rule:
• Total Volume of the new solid = Sum of the volumes of its individual parts.
• Total Surface Area (TSA) of the new solid = Sum of the Curved Surface Areas (CSA) of the visible individual parts. Do not add their base circles, because those surfaces are hidden inside the joint!

✏️ Complete NCERT Solutions Class 10 Surface Areas & Volumes

Exercise 12.1 (Rationalized Syllabus)
Q1. 2 cubes each of volume $64\text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.
Step 1: Find the edge side ($a$) of the initial cubes
$\text{Volume} = a^3 = 64 \implies a = \sqrt[3]{64} = 4\text{ cm}$
Step 2: Find dimensions of the newly joined cuboid
When placed side-by-side, only the length changes:
Length ($l$) = $4 + 4 = 8\text{ cm}$, Breadth ($b$) = $4\text{ cm}$, Height ($h$) = $4\text{ cm}$
Step 3: Calculate Surface Area of Cuboid
$\text{TSA} = 2(lb + bh + hl)$
$\text{TSA} = 2(8\times4 + 4\times4 + 4\times8) = 2(32 + 16 + 32)$
$\text{TSA} = 2(80) = \mathbf{160\text{ cm}^2}$
Final Answer: The surface area of the resulting cuboid is $160\text{ cm}^2$.