Question

A solid is in the shape of a frustum of a cone. The diameters of the two circular ends are 60 cm and 36 cm and the height id 9 cm. Find the area of its whole surface and the volume.

Answer 1

Answer:

Given height of a frustum cone = 9cm

Lower end radius (r1) = 60/2 = 30cm

Upper end radius (r2) = 36/2 = 18cm

Let slant height of frustum cone be l .

_______________

l = √(r1-r2)^2 + h^2

_____________

= √(8-30)^2 + 9 ^2

_________

= √(144+ 81

= 15 cm

Volume of frustum cone

=1/3 π (r1^2 +r2^2 + r1r2) h

= 1/3 π( 30^2 + 18^2 + 30×18) ×9

= 5292 π cm^3

Total surface area of frustum cone

= π(r1+r2) × l + πr1^2 + πr2^2

= π(30+ 18)15 + π(30)^2 + π ( 18)^2

= π( 48×15 + (30)^2 + (18)^2

= π(720+900+ 324 )

= 1944 π cm^2 .........ans.

or

1944 × 22/7 cm^2

= 42768/7 cm^2

= 6,109.714 cm^2

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Answer 2

Answer:

$\frac{1}{3} \pi h( R^{2} + r^{2} + Rr)$ = volume of frustum

$\frac{1}{3} X \frac{22}{7} X 9( 3600 + 324 + 540)= \frac{22}{7} X 3 ( 4464) = \frac{22}{7} X 13392 = 22 X 1913.14 = 42089.08 cm^{3}$

Total surface area of frustum = $\pi l( R + r) + \pi R^{2} + \pi r^{2}$

l = $\sqrt{h^{2} + (R - r)^{2} }$ = $\sqrt{81 + (12)^{2} } = \sqrt{81+144} = \sqrt{225} = 15$

$\frac{22}{7} X 15( 48) + \frac{22}{7} X 30 X 30 + \frac{22}{7} X 18 X 18 = \frac{22}{7} X 720 + \frac{22}{7} X 900 + \frac{22}{7} X 324 = \frac{22}{7} (720+ 900+ 324) = \frac{22}{7} ( 1944) = 22 X 277.71 = 6109.62 cm^{2}$

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