V(r,h)=⅓πr²h where V is the volume of a cone as a function of r, the base radius, and h, the vertical height of the cone. The volume of a frustum is the difference in volume of two similar cones.

The slant length l is related to r and h: l²=r²+h². Think of a pie with a piece taken out. If the pie has a radius of L and the angle of of the piece is θ radians, then the shape of the pie that’s left can be folded into a cone with a seam where the piece was taken out. The surface area, A, of the top of the piece is given by A/(πL²)=θ/2π that is, the ratio of the area of the piece to the area of the uncut pie (πL²) is the same as the ratio of the angle to 2π (equivalent to 360°). A=L²θ/2. The area of the remaining pie is πL²-L²θ/2=½L²(2π-θ). The circumference is shortened by Lθ, the sector length, to 2πL-Lθ=L(2π-θ), and this forms the base of the cone.

The base radius of the assembled cone is R so the circumference of the base is 2πR=L(2π-θ), from which θ can be found.

The surface area of a frustum will be the difference of the surface areas of two cones of different radii and slant lengths. Let R and r be the base radii of two similar cones and V and v be their volumes, then the volume of the frustum is V-v. The ratio of the heights (h/H) and slant lengths (l/L) of the cones is r/R, so if H is the height of the larger cone, rH/R is the height of the smaller cone. In the question, R=14cm and H-h=21cm.

The formulae we have so far:

V=⅓πR²H, H=3V/(πR²); v=⅓πr²h; V-v=⅓π(R²H-r²h); 2πR=L(2π-θ); 2πr=l(2π-θ); l/L=h/H=r/R; A=½L²(2π-θ) for the larger cone and a=½l²(2π-θ) for the smaller cone. We are given V-v=2401π cc; R=14cm assuming the lower base is larger; H-h=21cm; l²=h²+r²; L²=H²+R². So we can write: A=½(H²+R²)(2π-θ), a=½(h²+r²)(2π-θ) because the cone angles are the same, and a/A=(h²+r²)/(H²+R²)=(l/L)²=(h/H)²=(r/R)²=p², where p is the ratio r/R. We would expect this for similar figures. Similarly, v/V=p³ and V-v=V(1-p³).

From these: 2401π=⅓π(R²H-r²h)=⅓πR²H(1-p³), so R²H(1-p³)=7203.

H-h=H-pH=H(1-p)=21cm.

Since R=14, H(1-p³)=7203/196=147/4. But H(1-p³)=H(1-p)(1+p+p²)=21(1+p+p²).

Therefore 21(1+p+p²)=147/4, so 1+p+p²=7/4, p²+p-¾=0, 4p²+4p-3=0=(2p+3)(2p-1). Therefore p=½.

We now know r=14/2=7cm, H(1-p)=H/2=21 so H=42cm, h=21cm, L=√484+196=√680=2√170, l=√170.

2πr=l(2π-θ) so 14π=√170(2π-θ), 2π-θ=14π/√170.

a=½170×14π/√170=286.73 sq cm approx.

A=1146.92 (4 times as big) so SA=860.19 sq cm.