Question

The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area and the total suface area.​

Answer 1

$\huge{\textbf{\underline{\underline{ANSWER:}}}}$

Let 'ℓ' & R be the radii of the bottom and top circular ends of a frustum of a cone, respectively and 'h' be height of cone.

Volume of the frustum of the cone

$= \frac{1}{3} \times \frac{22}{7} \times 45 \times ( {28}^{2} + {7}^{2} + 28 + 28 \times 7) \: {cm}^{3} \\ = 330 \times 174 \: {cm}^{3}$

Now, l = √h² + (R - r²)

= √2466

= 45.65 cm

Curved surface area of the frustum of the cone

$\bold{\boxed{CSA= π(R + r)l}}$

= [ 22/7 × (28 + 7) × 49.65] cm²

= 22 × 5 × 49.65 cm²

= 5461.5 cm²

Total surface area of frustum of the cone

$\bold{\boxed{TSA= π(R + r)l + πR² + πr²}}$

= [5461.5+ 22/7 ×(28)² + 22/7 ×(7)²] cm²

= 8079.5cm²

Answer 2

Step-by-step explanation:

Let 'ℓ' & R be the radii of the bottom and top circular ends of a frustum of a cone, respectively and 'h' be height of cone.

Volume of the frustum of the cone

\begin{lgathered}= \frac{1}{3} \times \frac{22}{7} \times 45 \times ( {28}^{2} + {7}^{2} + 28 + 28 \times 7) \: {cm}^{3} \\ = 330 \times 174 \: {cm}^{3}\end{lgathered}

=

3

1

×

7

22

×45×(28

2

+7

2

+28+28×7)cm

3

=330×174cm

3

Now, l = √h² + (R - r²)

= √2466

= 45.65 cm

Curved surface area of the frustum of the cone

\bold{\boxed{CSA= π(R + r)l}}

CSA=π(R+r)l

= [ 22/7 × (28 + 7) × 49.65] cm²

= 22 × 5 × 49.65 cm²

= 5461.5 cm²

Total surface area of frustum of the cone

= [5461.5+ 22/7 ×(28)² + 22/7 ×(7)²] cm²

= 8079.5cm²