To show the development of the truncated cone, we first need to determine its key dimensions. Given from Figure (a): Base diameter of the cone = 60 mm Total height of the original cone = 100 mm Height of the truncated part (frustum) = 70 mm Step 1: 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, $$L = \sqrt{H^2 + R^2}$$ $$L = \sqrt{100^2 + 30^2}$$ $$L = \sqrt{10000 + 900}$$ $$L = \sqrt{10900} \approx 104.40 \text{ mm}$$ Step 2: 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, $l$: $$l = \sqrt{h^2 + r^2}$$ $$l = \sqrt{30^2 + 9^2}$$ $$l = \sqrt{900 + 81}$$ $$l = \sqrt{981} \approx 31.32 \text{ mm}$$ Step 3: 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: $$\theta = \frac{R}{L} \times 360^\circ$$ $$\theta = \frac{30}{104.40} \times 360^\circ$$ $$\theta \approx 0.287356 \times 360^\circ$$ $$\theta \approx 103.45^\circ$$ Step 4: Construct the development of the truncated cone (frustum). The development of the frustum is a portion of an annulus. 1. Draw