Answer:
(B) [tex]76\pi$ sq. in.[/tex]
Step-by-step explanation:
Since the base of the cone and one circular face of the cylinder is not visible,
The surface area of the rocket=Base Area of the Cylinder+Lateral area of the Cylinder+Lateral Area of the Cone+
[tex]R$adius =4 Inches\\Height of the cylinder =5 Inches\\Therefore:\text{Lateral area of a Cylinder+Base Area of the Cylinder}=2\pi rh+\pi r^2\\=(2\times \pi \times 4 \times 5)+(\pi \times 4^2)\\=40\pi+16\pi \\=56\pi$ sq. in.[/tex]
Lateral Area of a Cone [tex]=\pi rl[/tex]
- Base radius, r= 8/2 =4 Inches
- Perpendicular Height of the Cone = 3 Inches
Using Pythagoras theorem:
[tex]Hypotenuse =\sqrt{opposite^2+adjacent^2} \\=\sqrt{3^2+4^2} \\=\sqrt{25}\\ =5 in.[/tex]
- Slant Height of the Cone, l (Hypotenuse) = 5 Inches
Therefore: Lateral Area of a Cone [tex]=\pi \times 4\times 5 =20\pi$ sq. in.[/tex]
Therefore, Surface area of the rocket
[tex]=56\pi+20\pi\\=76\pi$ sq. in.[/tex]
