To show the development of the truncated cone, we first need to determine its key dimensions.

Question

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To show the development of the truncated cone, we first need to determine its key dimensions.

Given from Figure (a):

Determine the dimensions of the original full cone.
- Base radius, $R = \frac{60}{2} = 30$ mm
- Height, $H = 100$ mm
- Slant height, $L$ : Using the Pythagorean theorem, $L = \sqrt{H^2 + R^2}$ $L = \sqrt{100^2 + 30^2}$ $L = \sqrt{10000 + 900}$ $L = \sqrt{10900} \approx 104.40 \text{ mm}$
Determine the dimensions of the removed top cone.
The height of the removed top cone, $h = H - 70 = 100 - 70 = 30$ mm.
Using similar triangles, the radius of the top surface of the frustum (which is the base radius of the removed top cone), $r$ , can be found:
$\frac{r}{R} = \frac{h}{H}$ $r = R \times \frac{h}{H}$ $r = 30 \times \frac{30}{100}$ $r = 30 \times 0.3 = 9 \text{ mm}$
The slant height of the removed top cone, $l$ :
$l = \sqrt{h^2 + r^2}$ $l = \sqrt{30^2 + 9^2}$ $l = \sqrt{900 + 81}$ $l = \sqrt{981} \approx 31.32 \text{ mm}$
Calculate the sector angle for the development.
The development of the full cone is a sector of a circle with radius $L$ . The arc length of this sector is equal to the circumference of the base of the cone ( $2\pi R$ ).
The sector angle $\theta$ (in degrees) is given by:
$\theta = \frac{R}{L} \times 360^\circ$ $\theta = \frac{30}{104.40} \times 360^\circ$ $\theta \approx 0.287356 \times 360^\circ$ $\theta \approx 103.45^\circ$
Construct the development of the truncated cone (frustum).
The development of the frustum is a portion of an annulus.
1. Draw

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