Can we find out the area of conical frustum by using triangles?

Question

I have been trying to find out the area of conical frustum by using triangles.

Answer 1

By area I assume you mean surface area. Let s = slant height; R = bigger radius (bottom radius); r = smaller radius (top radius); h = vertical height.

We can figure out the slant height by using the Pythagorean Theorem. Drop down the height from the upper circle, from where the r meets the edge of circle.

The base of the triangle is then R - r and the height is h, giving us that s is:

$s = \sqrt{h^2+(R-r)^2}$

Deriving the lateral surface area of a conical frustum makes use of similar triangles.

In this diagram L is used for slant height instead of s.

Since the right triangle formed by L, h and R-r is similar to the right triangle made by $L_1$, R and the total height of the cone, we can say that corresponding sides are in proportion and thus:

$\frac{L_1}{R} = \frac{L}{R-r}$

or $L_1 = \frac{RL}{R-r}$

From the figure: $ L_2 = L_1 - L $

$ L_2 = \dfrac{RL}{R - r} - L $

$ L_2 = \dfrac{RL - (R - r)L}{R - r} $

$ L_2 = \dfrac{rL}{R - r} $

The length of arc is the circumference of the base:

$ s_1 = 2\pi R $

$ s_2 = 2\pi r $

Since the lateral surface area of the frustum is the area of the side "unwound", which is the difference of the areas of sectors of a circle of radii $L_1$ and $L_2$ and common central angle θ, we can figure out the area by subtracting the area of the two sectors. The area of a sector with radius q and arc-length P is given by

$A = \pi q^2 \cdot \frac{P}{2\pi q} = \frac{q \cdot P}{2}$.

Returning back to our problem, we can get (see the figure) the lateral surface area by subtracting the area of the smaller sector from the larger: $ L.S.A = \frac{1}{2}s_1 \, L_1 - \frac{1}{2}s_2 \, L_2 $

$ L.S.A = \frac{1}{2}(2\pi R) \left( \dfrac{RL}{R - r} \right)- \frac{1}{2}(2\pi r) \left( \dfrac{rL}{R - r} \right) $

$ L.S.A = \dfrac{\pi R^2 L}{R - r} - \dfrac{\pi r^2 L}{R - r} $

$ L.S.A = \dfrac{\pi R^2 L - \pi r^2 L}{R - r} $

$ L.S.A = \dfrac{\pi (R^2 - r^2) \, L}{R - r} $

$ L.S.A = \dfrac{\pi (R - r)(R + r) \, L}{R - r} $

$ L.S.A = \pi(R + r)L $

(The source for this proof is: http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-lateral-area-of-a-right-circular-cone )

Returning to our original formula, we can substitute in for L = s, giving the formula for the lateral surface area:

L.S.A. = $\pi(R+r)(\sqrt{h^2+(R-r)^2})$

To get the total surface area we just add the surface area of the bases:

T.S.A = $\pi(R+r)(\sqrt{h^2+(R-r)^2}) + \pi(R^2+r^2)$