. In this exercise you will prove that the surface area of a frustum with radii p and q and slant length s is equal to SA = 2πrs, where $r=\frac{p+q}{2}$is the average radius of the frustum. The argument will hinge on the fact that a frustum is a cone with its top removed, as shown here:

(a) Use similar triangles to prove that, in the notation of the given diagram, pt + ps = qt.

(b) We know from Theorem 3.12 that the surface area of a cone with radius A and height B is given by $\mathrm{\pi A}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}$. Use the Pythagorean theorem to prove that this expression implies that the surface area of a cone with radius A and slant height C is given by πAC.

(c) Express the surface area of the frustum in the figure as the difference of the area of the larger cone and the area of the smaller cone, and use the previous two parts of this problem to prove that the surface area of the frustum is SA = π(p + q)s.

(d) Finally, use the relationship $r=\frac{p+q}{2}$to prove that the surface area of the frustum is S = 2πrs.

The surface area of a frustum with radii p and q and slant length s is equal to SA = 2πrs, where $r=\frac{p+q}{2}$is the average radius of the frustum

Observe that the triangle APQ and ABC are similar triangle . So the ratio of like sides of the triangle will be equal . That is

$$\begin{gathered} \frac{PQ}{BC}=\frac{AQ}{AC} \\ \frac{p}{q}=\frac{t}{s+t} \\ p(s+t)=qt \\ \mathrm{Hence}\mathrm{proved}pt+ps=qt \end{gathered}$$

Remember that in a right angled triangle the slant height or hypotenuse is the square root of the sum of squares of base and perpendicular. That is in the right triangle PQR given below

$C=\sqrt{A^2+B^2}$

Substitute this value of $\sqrt{A^2+B^2}$in the expression for surface area to obtain that the surface area of a cone with radius A and slant height C is given by $\mathrm{\pi }$AC

The surface area of a cone of radius rand slant height s is given by SA=2$\mathrm{\pi }$rs. Use this and the result of part (a) to write the area of frustum of the cone given in figure 1 as

SA = Surface area of the larger cone - surface area of the smaller cone

$$\begin{gathered} SA=\mathrm{\pi q}(\mathrm{s}+\mathrm{t})-\mathrm{\pi pt} \\ =\mathrm{\pi }\left(\mathrm{qs}+\mathrm{qt}-\mathrm{pt}\right) \\ =\mathrm{\pi }\left(\mathrm{qs}+\mathrm{ps}\right) \\ =\mathrm{\pi }(\mathrm{p}+\mathrm{q})\mathrm{s} \end{gathered}$$

Use the relation $r=\frac{p+q}{2}$in the expression for surface area of the frustum

obtained in part (c) to get

S = n(2r)s

=$2\mathrm{\pi rs}$

Therefore, the surface area of the frustum of the cone is$2\mathrm{\pi rs}$