Question -
Answer -
Given,
Perimeter of the upperend = 44 cm
2 π r1 =44
2(22/7) r1 =44
r1 = 7cm
Perimeter of the lowerend = 33 cm
2 π r2 =33
2(22/7) r2 =33
r2 =21/4 cm
Now,
Let the slant heightof the frustum of a right circular cone be L
L = 16.1 cm
So, the curved surfacearea of the frustum cone = π(r1 + r2)l
= π(7 + 5.25)16.1
Curved surface area ofthe frustum cone = 619.65 cm3
Next,
The volume of thefrustum cone = 1/3 π(r22 + r12 +r1 r2 )h
= 1/3 π(72 +5.252 + (7) (5.25) ) x 16
= 1898.56 cm3
Thus, volume of thecone = 1898.56 cm3
Finally, the totalsurface area of the frustum cone
= π(r1 +r2) x L + π r12 + π r22
= π(7 + 5.25) × 16.1 +π72 + π5.252
= π(7 + 5.25) ×16.1 + π(72 + 5.252) = 860.27 cm2
Therefore, the totalsurface area of the frustum cone is 860.27 cm2