Question

the sum of ends of frustum of cones is 45 cm high and 28 cm and 7 cm. find the volume, the csa and tsa​

Answer 1

Mistake in question:

Correct question :

The radii of the ends of frustum of cone is 45 cm high and 28 cm and 7 cm. Find the volume, CSA and TSA.

Answer:

$\large \text{volume of frustum =48,510 \ cm^3 }$

$\large \text{CSA of frustum =5,462.38 \ cm^2 }$

$\large \text{TSA of frustum =8,080.38 \ cm^2 }$

Step-by-step explanation:

Given :

Height of  frustum = 45 cm  and radii = 28 cm & 7 cm

We know formula of volume of frustum :

$\large \text{volume of frustum =\dfrac{1}{3}\pi h (R^2+r^2+Rr) }$

where h = height of  frustum and R & r are radii of  frustum.

Putting values here we get

$\large \text{volume of frustum =\dfrac{1}{3}\pi \times45 (28^2+7^2+28\times7) \ cm^3 }\\\\\\\large \text{volume of frustum = \dfrac{22}{7}\times15 (28^2+7^2+28\times7) \ cm^3 }\\\\\\\large \text{volume of frustum = \dfrac{22}{7}\times15 (784+49+196) \ cm^3 }\\\\\\\large \text{volume of frustum = \dfrac{22}{7}\times15 (1029) \ cm^3 }\\\\\\\large \text{volume of frustum =22\times15\times147 \ cm^3 }\\\\\\\large \text{volume of frustum =48,510 \ cm^3 }$

Now for CSA ( Curved surface area )

We have formula

$\large \text{CSA of frustum =\pi (R+r) l \ cm^2 }$

Here we have to find l ( slant height )

$\large \text{ l=\sqrt{h^2+(R-r)^2} put value here}\\\\\\\large l=\sqrt{45^2+(28-7)^2} \\\\\\\large l=\sqrt{45^2+(21)^2} \\\\\\\large l=\sqrt{ 2025+441} \\\\\\\large l=\sqrt{ 2466} \\\\\\\large l=49.658 \ cm$

Putting in CSA formula we get

$\large \text{CSA of frustum =\pi (R+r) l \ cm^2 }\\\\\\\large \text{CSA of frustum =\dfrac{22}{7}(28+7)\times49.658 \ cm^2 }\\\\\\\large \text{CSA of frustum =\dfrac{22}{7}(35) \times49.658 \ cm^2 }\\\\\\\large \text{CSA of frustum =22\times5\times49.658 \ cm^2 }\\\\\\\large \text{CSA of frustum =5,462.38 \ cm^2 }$

Now TSA ( Total surface area )

We have formula

$\large \text{TSA of frustum =\pi (R+r) l+\pi R^2+\pi r^2 \ cm^2 }$

putting values here we get

$\large \text{TSA of frustum =5,462.38+\pi( 28^2+ 7^2) \ cm^2 }\\\\\\\large \text{TSA of frustum =5,462.38+\pi( 784+49) \ cm^2 }\\\\\\\large \text{TSA of frustum =5,462.38+\dfrac{22}{7} ( 833) \ cm^2 }\\\\\\\large \text{TSA of frustum =5,462.38+ 22\times119 \ cm^2 }\\\\\\\large \text{TSA of frustum =8,080.38 \ cm^2 }$

Thus we get all answer.

Answer 2

Correction in the question : The radii of ends of frustum of cones is 45 cm high and 28 cm and 7 cm. find the volume, the csa and tsa.

GIVEN :

Height of the frustrum (h) = 45cm

Let 'R' be the radius of top circular end.

'r' be the radius of bottom circular end.

R = 28cm

r = 7cm

Slant Height of the frustrum :

$= \sqrt{ {(h)}^{2} + {(larger \: radius - small \: radius)}^{2} } \\ \\ = \sqrt{ {(45)}^{2} + {(28 - 7)}^{2} } \\ \\ = \sqrt{ {(45)}^{2} + {(21)}^{2} } \\ \\ = \sqrt{2025 + 441} \\ \\ = \sqrt{2466} \\ \\ = 49.65$

Volume of Frustrum = 1/2πh [ R² + r² + Rr]

= 1/3 × 22/7 × 45 [ (28)² + (7)² + 28 × 7 ]

= 1/3 × 22/7 × 45 [ 784 + 49 + 196 ]

= 22/7 × 15 [ 1029 ]

= 330 [ 147 ]

= 48510cm³

TSA of Frustrum = π(R + r)l + πR² + πr²

= 22/7(28 + 7)49.65 + (22/7 × (28)² ) + 22/7(7)²

= 22/7 (35)49.65 + (22/7 × 784) + (22/7(49)

= 22(5)49.65 + (22 × 112) + (22 × 7)

= 110 × 49.65 + 2464 + 154

= 5461.5 + 2464 + 154

= 8079.5cm²

CSA of Frustrum = π(R + r)l

= 22/7(28 + 7)49.65

= 22/7(35) × 49.65

= 22 × 5 × 49.65

= 110 × 49.65

= 5461.5cm²