Right Circular Cone

A Right Circular Cone is a three-dimensional geometric solid formed by rotating a right-angled triangle through one complete revolution ($360^\circ$) about one of the sides containing the right angle, keeping that side fixed.

• Apex or Vertex (শীর্ষবিন্দু): The fixed top point $A$ of the cone opposite to the circular base.
• Axis (অক্ষ): The straight line segment joining the apex $A$ to the center of the base $O$. In a right circular cone, this axis is strictly perpendicular to the plane of the circular base.
• Circular Base (বৃত্তাকার ভূমি): The flat circular bottom surface traced by the perpendicular leg of the rotating right triangle.
• Base Radius (ভূমির ব্যাসার্ধ, $r$): The radius $OB$ of the circular base.
• Vertical Height (উচ্চতা, $h$): The perpendicular distance from the apex $A$ to the center of the circular base $O$.
• Slant Height (তির্যক উচ্চতা, $l$): The distance from the apex $A$ to any point $B$ on the circumference of the base circle, generated by the hypotenuse of the rotating right triangle.

It is termed circular because its base is a perfect circle. It is termed right because the straight line joining the vertex to the center of the base meets the base at a right angle ($90^\circ$). If the apex is tilted such that the axis is not perpendicular to the base, it is an oblique cone (which is outside the WBBSE secondary syllabus).

In the right circular cone, triangle $ riangle AOB$ is a right-angled triangle with $ngle AOB = 90^\circ$. By Pythagoras' Theorem:

From this fundamental relationship, if any two of the three parameters are known, the third can be found immediately:

• To find Slant Height: $\mathbf{l = \sqrt{r^2 + h^2}}$
• To find Vertical Height: $\mathbf{h = \sqrt{l^2 - r^2}}$
• To find Base Radius: $\mathbf{r = \sqrt{l^2 - h^2}}$

Madhyamik examination questions frequently use standard integer Pythagorean triplets $(r, h, l)$:

• $(3, 4, 5)$ or multiples: $(6, 8, 10)$, $(9, 12, 15)$, $(15, 20, 25)$
• $(5, 12, 13)$ or multiples: $(10, 24, 26)$
• $(7, 24, 25)$
• $(8, 15, 17)$

To find the area of the curved surface, imagine cutting the cone open along a slant edge $AB$ and unrolling it flat on a table. The resulting shape is a sector of a circle:

• The radius of this large sector is the slant height $l$ of the cone.
• The curved arc length of the sector is equal to the perimeter of the circular base of the cone: $ ext{Arc length} = 2\pi r$.

Now, from the geometry of a circular sector: $$ ext{Area of Sector} = rac{1}{2} imes ext{Arc Length} imes ext{Radius of Sector}$$ $$ ext{CSA} = rac{1}{2} imes (2\pi r) imes l = \mathbf{\pi r l}$$

The central angle $ heta$ (in degrees) of the unfolded sector is given by: $$rac{ heta}{360^\circ} = rac{ ext{Arc Length}}{ ext{Total Circumference of Radius } l} = rac{2\pi r}{2\pi l} = rac{r}{l}$$ $$\mathbf{ heta = rac{r}{l} imes 360^\circ}$$

A closed right circular cone consists of two distinct surfaces: the curved lateral surface and the flat circular base.

The volume (or capacity) of a right circular cone of base radius $r$ and vertical height $h$ is given by:

Consider a hollow cylinder and a hollow cone having the exact same base radius $r$ and height $h$. If you fill the cone completely with fine sand or water and pour it into the cylinder, it takes exactly three full cones to fill the cylinder completely!

The ratio of volumes of a cone and a cylinder having equal base and equal height is strictly $\mathbf{1 : 3}$.

When a conical tent is pitched on the ground:

• The floor area occupied by people or equipment is the flat circular base: $\mathbf{ ext{Floor Area} = \pi r^2}$.
• The canvas (তিরপল বা কাপড়ের ক্ষেত্রফল) required to cover the tent is strictly the Curved Surface Area ($\mathbf{ ext{Canvas} = \pi r l}$), because the ground/floor is NOT made of tent canvas!
• The air volume inside the tent available for breathing is the total volume: $\mathbf{V = rac{1}{3}\pi r^2 h}$.
• If $N$ persons sleep inside the tent, and each person requires $a ext{ m}^2$ of floor space and $v ext{ m}^3$ of air: $$\pi r^2 = N imes a \quad ext{and} \quad rac{1}{3}\pi r^2 h = N imes v$$ $$\implies rac{1}{3}(N imes a) imes h = N imes v \implies \mathbf{h = rac{3v}{a}}$$ The required height of the tent is independent of the number of persons $N$!

When a cylindrical bucket of sand or grain is emptied onto the ground, it naturally forms a conical heap due to the friction between particles. By the Principle of Conservation of Volume: $$ ext{Volume of Cylindrical Container} = ext{Volume of Conical Heap}$$ $$\pi R_{ ext{cyl}}^2 H_{ ext{cyl}} = rac{1}{3}\pi r_{ ext{cone}}^2 h_{ ext{cone}}$$