Question

A right circular cone of height h and radius R is cut by a plane parallel to the base at a distance h/5 from the base. The ratio of the curved surface area of the frustum and the total surface area of the resulting cone is:

• A. 9l : 16(l + R/5)
• B. 9l : 5(l + R)
• C. 16l : 9(l + R)
• D. 9l : 16(l + R)

Answer 1

Answer:

D. 9l : 16(l + R)

A cone of height (h) and radius (R) is cut at a distance (h/5) from the base.

Dimensions of the smaller cone

Height of small cone:
$h - \frac{h}{5} = \frac{4h}{5}$

By similarity:
$\frac{r}{R} = \frac{4h/5}{h} = \frac{4}{5}$
$\Rightarrow r = \frac{4R}{5}$
Slant heights

Original cone slant height = (l)

Small cone slant height:
$l_s = \frac{4}{5}l$
Curved surface area (CSA) of frustum

$\text{CSA of frustum} = \pi (R + r)(l - l_s)$
$= \pi \left(R + \frac{4R}{5}\right)\left(l - \frac{4l}{5}\right)$
$= \pi \left(\frac{9R}{5}\right)\left(\frac{l}{5}\right)$
$= \frac{9}{25}\pi Rl$
Total surface area (TSA) of small cone

$\text{TSA} = \pi r l_s + \pi r^2$
$= \pi \left(\frac{4R}{5}\cdot \frac{4l}{5}\right) + \pi \left(\frac{4R}{5}\right)^2$
$= \frac{16}{25}\pi Rl + \frac{16}{25}\pi R^2$

Ratio

$\frac{\text{CSA of frustum}}{\text{TSA of small cone}}$
$= \frac{\frac{9}{25}\pi Rl}{\frac{16}{25}\pi (Rl + R^2)}$

Cancel common terms:
$= \frac{9Rl}{16(Rl + R^2)} = \frac{9l}{16(l + R)}$
Final Answer:

$\boxed{9l : 16(l + R)}$