Question

The radii of circular ends a solid frustum of a cone are 33cm and 27cm and its slant height is 10cm Find its capacity and its total surface area.Take π=22/7

urgent please solve fast

Answer 1

radii of solid circular frustum

r = 27cm , R = 33cm

slant height l = 10cm

height h^2 = l^2 - ( R - r)^2

h^2 = (10)^2 - (33 - 27)^2

h^2 = 100 - 36 = 64

h = √64 = 8cm

height of solid circular frustum= 8cm

capacity = volume of solid circular frustum

volume = (πh/ 3)*( R^2 + Rr + r^2)

= (22×8/7×3) * [ (33)^2 + 33 × 27 + (27)^2]

= 22704cm^3

total surface area

= π( r + R) × l + π( r^2 + R^2)

=[ 22(27 + 33)× 10)/ 7 ]+22[ (27)^2+ (33)^2]/ 7

= (13200/ 7) + (39996/ 7)

= 7599.42 cm^2

Answer:
---------------

capacity = 22704cm^3,

total surface area = 7599.42cm^2

Answer 2

$\huge{\bold{\fbox{\underline{\mathbb{SOLUTION:-}}}}}$

Let R and r be the radio of the top and the base of the bucket respectively , and let him be It heights.

Then,

$\bf{GIVEN:-}$

R = 33 Centimeters

r = 27 Centimeters

l = 10 Centimeters

•°• H =
$\mathsf{ \huge{ \bf{ \sqrt{ {l}^{2} - {(R - r)}^{2}cm }}}} \\ \\ \\ \implies \mathsf{ \huge{ \bf{ \sqrt{ {10}^{2} - {(33 - 27)}^{2}cm }}}} \\ \\ \\ \\ \implies{ \huge{ \bf{ \sqrt{ {(10)}^{2} - {(6)}^{2} cm}}}} \\ \\ \\ \implies{ \huge{ \bf \: \sqrt{64 \: cm}}} \\ \\ \\ \implies{ \huge{ \bf{8 \: cm}}}$

$<tt>$

Now, According to the Question Statement!!

Capacity of the frustum = 1/3πh(R²+r²+Rr)

Substitute the Given value in Formula Equation!

Capacity of the frustum = 1/3πh(R²+r²+Rr)

=> 1/3×22/7×8[(33)²+(27)²+33(27)

=> 1/3 × 68112

=> 22704.00 cm³

Again,

Total Surface Area = (πR²+πr²+πl(R+r)

= π(R²+r²+l(R+r)

= 22/7 [(33)² + (27)²+10(33+27)

= 22/7 [1089+729 + 10(60)]

= 22/7 [ 1089+729+600]

= 22/7 × 2418

= 22 × 345.42

= 7599.43 cm³

Hence, Required its capacity and its total surface area are respectively 22704 cm³and 7599.43 cm³.