Question

The radii of the ends of a frustum of a cone 45 cm high are 28cm and 7cm. Find its volume, the curved surface area. [Take π=22/7]

Answer 1

Given :-

• The radii of the ends of a frustum of a cone 45 cm high are 28cm and 7cm

To find :-

• Volume of frustum
• Curved surface area of frustum

Solution :-

Let the height be h , slant height be l and its radius be R and r where R > r

• Height of frustum (h) = 45cm

• Radii of frustum (R and r) = 28cm and 7cm

As we know that

→ Volume of frustum = ⅓ πh(R² + r² + Rr)

Substitute all the values of height and radii

→ ⅓ πh(R² + r² + Rr)

→ ⅓ × 22/7 × 45[(28)² + (7)² + 28 × 7]

→ 990/21[784 + 49 + 196]

→ 990/21[784 + 245]

→ 990/21 × 1029

→ 990 × 49

→ 48510cm³

Hence,

• Volume of frustum is 48510cm³

Now, curved surface area of frustum

Firstly, we have to find out slant height (l) of the frustum.

For calculating slant height (l)

→ l = √h² + (R - r)²

→ l = √(45)² + (28 - 7)²

→ l = √2025 + (21)²

→ l = √2025 + 441

→ l = √2466

→ l = 49.65cm

Curved surface area of frustum = π(R + r)l

→ π(R + r)l

Put the value of radii and slant height

→ 22/7(28 + 7) × 49.65

→ 22/7 × 35 × 49.65

→ 22 × 5 × 49.65

→ 110 × 49.65

→ 5461.5 cm²

Hence,

• Curved surface area of frustum is 5461.5cm²

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The radii of the ends of a frustum of a cone 45 cm high are 28cm and 7cm.

$\bf{\underline{\underline\blue{To\: Find:-}}}$

• The volume of frustum.

• The curved surface area of the frustum.

$\bf{\underline{\underline\green{Solution:-}}}$

As we know that

Volume of frustum is given by the formula:-

$\boxed{\rm Volume \: of \: frustum = \frac{1}{3} \pi h( { r_{1}}^{2} + { r_{2} }^{2} + r_{1} r_{2} ) \: }$

Here:-

• h = Height = 45cm

• $\rm r_{1} = 28cm$

• $\rm r_{2} = 7cm$

Substituting the values:-

$= {\rm \dfrac{1}{3} \times \dfrac{22}{7} \times 45 \bigg[ ( {28)}^{2} + { (7) }^{2} + (28)(7) \bigg ]\: }$

$= {\rm \dfrac{1}{3} \times \dfrac{22}{7} \times 45 \bigg[ 784+ 49 + 196\bigg ]\: }$

$= {\rm \dfrac{1}{3} \times \dfrac{22}{7} \times 45 \times 1029}$

$= {\rm \dfrac{1}{ \cancel3} \times \dfrac{22}{ \cancel7} \times \cancel{45} \: \large^{15} \times \cancel{1029} \: \: \large^{147} }$

$= {\rm22 \times 15 \times 147 }$

$= {\rm48510 }$

$\large\underline{ \boxed{\rm\pink{ Volume \: of \: frustum = 48510 {cm}^{3} }}}$

Now:-

$\implies{\rm l = \sqrt{ {h }^{2} + ( { r_{1} } - { r_{2})}^{2} } }$

Substituting the values:-

$\implies{\rm l = \sqrt{ {(45) }^{2} + ( { 28 } - { 7)}^{2} } }$

$\implies{\rm l = \sqrt{ { 2025} + ( { 21)}^{2} } }$

$\implies{\rm l = \sqrt{ { 2025} + 441} }$

$\implies{\rm l = \sqrt{ 2466} }$

$\implies{\rm \underline{\underline { \red{ \: \: l = 49.65cm \: \:} }}}$

Now:-

Curved surface area (CSA) of frustum is given by the formula:-

$\boxed{\rm{ \leadsto CSA \: of \: frustum = \pi( r_{1} + r_{2})l }}$

Substituting the values

${\rm{ = \dfrac{22}{7} \times \big( 28 + 7\big) \times 49.65}}$

${\rm{ = \dfrac{22}{ \cancel7} \times \cancel{35 } \: \: \large^{5} \times 49.65}}$

${\rm{ = 22 \times 5 \times 49.65}}$

${\rm{ = 5461.5}}$

$\large\underline{\boxed{\rm \purple{ CSA \: of \: frustum = 5461.5 {cm}^{2} }}}$