Question

In the given figure, ABCD is a trapezium. The curved surface area of the solid obtained when given trapezium is revolved about AD is

• A. $2024c{m}^2$
• B. $1048c{m}^2$
• C. $1470c{m}^2$
• D. $1650c{m}^2$

Answer 1

The correct option is D $1650c{m}^2$

If we revolve the given trapezium about AD, then the solid formed is a cylinder surmounted by a cone.

∴ Radius of cone (r) = DC = 7cm = Radius of cylinder (r)

Height of cylinder (h) = BC = 25 cm

Height of cone (h′) = AD – BC = 49 – 25 = 24 cm



$\therefore$ Slant height of cone (l) $=\sqrt{{24}^2+7^2}=25cm$

$\therefore$ CSA of solid = CSA of cylinder + CSA of cone

$=2\pi rh+\pi rl$

$=2\pi \times 7\times 25+\pi \times 7\times 25$

$=7\pi (50+25)$

$=7\times \frac{22}{7}\times 75$

$=1650c{m}^2$

Hence, the correct answer is option (d).