Question

The perimeters of the ends of the frustum of a cone are 96 cm and 68 cm. If the height of the frustum be 20 cm, find its radii, slant height,and total surface.​

Answer 1

Answer:

Total surface equal to $= 2780.89 cm^{2}$

Step-by-step explanation:

Let the larger and smaller radii of the frustum be R and r respectively.

Perimeters of the end of the frustum are 96 cm and 68 cm.

Therefore:

• 2πR = 96 and 2πr = 68

$2 * \frac{22}{7} * R = 96$

$R = \frac{96 * 7}{2 * 22} = 15.27 cm$

• 2πr = 68

$2 * \frac{22}{7} * r = 68$

$r = \frac{68 * 7}{2 * 22} = 10.82 cm$

Let h be the height of the frustum, h = 20 cm,

Slant height, l, is equal to:

$\sqrt{(R - r)^{2} + h^{2}} = \sqrt{(15.27 - 10.82)^{2} + 20^{2}} = \sqrt{4.45^{2} + 20^{2} } = \sqrt{419.8025}$

$l = 20.49 cm$

Total surface area of frustum:

$= pi(R + r)l + piR^{2} + pir^{2}$

$=\frac{22}{7} * (15.27 + 10.82) * 20.49 + \frac{22}{7} * (15.27^{2} + 10.82^{2} )$

$=\frac{22}{7} * 26.09 * 20.49 + \frac{22}{7} * (233.1729 + 117.0724)$

$= 1680.12 + \frac{22}{7} * 350.2453$

$= 1680.12 + 1100.77$

$= 2780.89 cm^{2}$