Question

the perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56cm.if the height of the frustum be 8 cm,find the whole surface area of the frustum. (use .3.14)

Answer 1

Perimeter of 1st end = 2 π R 

⇒ 207. 24 cm = 2 x 3.14 x R 
⇒ 207.24 cm =  6.28 cm x R
⇒  33 = R
⇒ i .e R = 33 

Perimeter or 2nd end = 2 π r
⇒ 169. 56 cm = 2 x 3.14 x r
⇒ 169. 56 cm  = 6.28 cm x r
⇒ 27 cm = r 
⇒ i .e. , r = 27 cm

So ,
We need to find slant height first .

R = 33 cm 
r = 27 cm
H = 8 cm 
 
⇒ slant height =  $\sqrt{( (r _{1} - r_{2} }) ^{2} + \sqrt{h ^{2} }$

⇒ Slant height = $\sqrt{( (33 - 27) ^{2} + {8 ^{2} }$

⇒ Slant height = $\sqrt{( 360 ) + {64 }$

⇒ Slant height = 20 .59 , round it to 20.6 cm

⇒Total surface area = π ( r + R ) l + π R² + π r ²

⇒ Total surface area = 3.14 ( 27 + 33 ) 20.6 + 3.14 ( 33)² + 3.14(27)²

⇒ Total surface area = 3.14 ( 60 ) (20.6) + 3.14 ( 1089 ) + 3.14 ( 729)

⇒ Total Surface area =  3881.04 + 3419.46 + 2289.06

⇒ Total Surface area = 9589.56 cm²

Answer 2

Answer:

 = 7592.52 cm²

Step-by-step explanation:

Let R and r be the radii of the circular ends of the frustum. (R> r) 

2R = 207.24 

R = 207.24/ (2 X 3.14)

 R = 33 cm

2r = 169.56 cm 

r = 169.56 / (2 X 3.14) 

r = 27 cm

1² = h² + (R-r)² 

= 8² + (33-27)² 

l = 10 cm  

Whole surface area of the frustum

 =  (R2 + r2 + (R+r)l)

 = 3.14

((33)2 + (27)2 + (33+27)10)

= 3.14 (1089 +729 + 600)

 = 3.14 X 2418 cm² 

 = 7592.52 cm²