Question

There is a toy rocket in the shape shown below. The base radius of the cylinder = 1.5 cm. Base radius of the cone = 2.5 cm. The total surface area (in sq.cm) of the toy .

• A. 82.5$\pi$
• B. 70.00$\pi$
• C. 90.5$\pi$

Answer 1

The correct option is A 82.5$\pi$

By observing the toy it can be seen that the following surfaces will be visible for us to see-

The curved surface of the cone

The curved surface of the cylinder

Base surface of the cylinder

Circular surface at the junction of the cylinder and cone

When we look at the toy from a top view, we can see the following.

When the cone is placed on top of the cylinder the region that is common between cylinder and cone is the region marked white. The remaining region which is marked blue is not overlapping and will hence be visible.

The area of blue region = Area of larger circle – Area of smaller circle

= $\pi$$({2.5}^2-{1.5}^2)$

= $\pi$(6.25- 2.25)

= 4$\pi$

C.S.A of cone = $\pi$rl = $\pi$r$\sqrt{(r^2+h^2)}$

= $\pi$(2.5)$\sqrt{({2.5}^2+6^2)}$

= $\pi$(2.5)(6.5)

= 16.25$\pi$ $c{m}^2$

Height of cylinder = Total toy height – Cone height

= 26 – 6

h = 20 cm

C.S.A of cylinder = 2$\pi$rh

= 2$\pi$(1.5)(20)

= 60$\pi$ $c{m}^2$

Base area of cylinder = $\pi$$r^2$

= $\pi$$(1.5{)}^2$

= $2.25$ $\pi$

Total surface area of toy = Sum of surface areas of all surfaces
$=(60+2.25+16.25+4)\pi$
$=82.5\pi$