Edexcel Mathematics (4XMAH) -Unit 2 - 4.10 3D Shapes and Volume- Study Notes- New Syllabus
Edexcel Mathematics (4XMAH) -Unit 2 – 4.10 3D Shapes and Volume- Study Notes- New syllabus
Edexcel Mathematics (4XMAH) -Unit 2 – 4.10 3D Shapes and Volume- Study Notes -Edexcel iGCSE Mathematics – per latest Syllabus.
Key Concepts:
A find the surface area and volume of a sphere and a right circular cone using relevant formulae
Edexcel iGCSE Mathematics -Concise Summary Notes- All Topics
Surface Area and Volume of Spheres and Cones
In mensuration we calculate how much space a 3D object occupies (volume) and how much material covers it (surface area).
Sphere
A sphere is a perfectly round 3D shape where every point on the surface is the same distance from the centre.
Let the radius be \( r \).
Surface Area of a Sphere
\( A=4\pi r^2 \)
Volume of a Sphere
\( V=\dfrac{4}{3}\pi r^3 \)
Right Circular Cone
A cone has a circular base and a single point (vertex) directly above the centre of the base.
Let:
- radius \( r \)
- height \( h \)
- slant height \( l \)
The slant height is found using Pythagoras:
\( l=\sqrt{r^2+h^2} \)
Curved Surface Area of a Cone
\( A=\pi rl \)
Total Surface Area of a Cone
\( A=\pi rl+\pi r^2 \)
Volume of a Cone
\( V=\dfrac{1}{3}\pi r^2 h \)
Example 1:
Find the volume of a sphere with radius \( 3 \) cm.
▶️ Answer/Explanation
\( V=\dfrac{4}{3}\pi r^3 \)
\( V=\dfrac{4}{3}\pi(3^3) \)
\( V=\dfrac{4}{3}\pi(27)=36\pi \)
Conclusion: \( 36\pi\text{ cm}^3 \).
Example 2:
Find the surface area of a sphere with radius \( 5 \) cm.
▶️ Answer/Explanation
\( A=4\pi r^2 \)
\( A=4\pi(5^2)=4\pi(25)=100\pi \)
Conclusion: \( 100\pi\text{ cm}^2 \).
Example 3:
A cone has radius \( 4 \) cm and height \( 3 \) cm. Find its volume.
▶️ Answer/Explanation
\( V=\dfrac{1}{3}\pi r^2h \)
\( V=\dfrac{1}{3}\pi(4^2)(3) \)
\( V=\dfrac{1}{3}\pi(16)(3)=16\pi \)
Conclusion: \( 16\pi\text{ cm}^3 \).