Question

the slant height of a frustum of a cone is 8cm and the circumstances of its circular ends are 28cm and 14cm. find the curved surface area of the frustum​

Answer 1

Appropriate Question :-

The slant height of a frustum of a cone is 8cm and the circumference of its circular ends are 28cm and 14cm. Find the curved surface area of the frustum.

$\large\underline{\sf{Solution-}}$

Given that,

Slant height of a frustum of a cone = 8cm.

Circumference of its circular ends are 28cm and 14cm.

Let assume that the radius of circular ends having perimeter 28 cm and 14 cm be R cm and r cm respectively.

So, we have

$\rm \: 2\pi \:R = 28 \: \: \rm\implies \:R = \frac{14}{\pi} \: cm$

and

$\rm \: 2\pi \:r = 14 \: \: \rm\implies \:r = \frac{7}{\pi} \: cm$

So, we have now

$\rm \: R = \frac{14}{\pi} \: cm$

$\rm \: r = \frac{7}{\pi} \: cm$

$l \: = \: 8 \: \rm \: cm$

Now, we know

Curved Surface Area of frustum of radius of its circular ends r and R respectively and slant height l is given by

$\boxed{{  \: \: CSA_{(Frustum)} = \pi(r + R)l \: \: }}$

So, on substituting the values, we get

$\rm \: CSA_{(Frustum)} = \pi\bigg(\dfrac{14}{\pi} + \dfrac{7}{\pi} \bigg) \times 8$

$\rm \: CSA_{(Frustum)} = (14 + 7) \times 8$

$\rm \: CSA_{(Frustum)} = 21 \times 8$

$\rm\implies \:\rm \: CSA_{(Frustum)} = 168 \: {cm}^{2}$

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Additional Information :-

$\boxed{\sf{  \:\rm \: Volume_{(Frustum)} = \frac{\pi \: h}{3} \: ( {R}^{2} + Rr + {r}^{2}) \: \: }}$

$\boxed{{  \:\ \: TSA_{(Frustum)} = \pi\bigg( {R}^{2} + {r}^{2} + (R + r)l \bigg) \: \: }}$

$where \: l \: = \rm \: \sqrt{ {(R - r)}^{2} + {h}^{2} }$

Answer 2

Answer:

168 cm²

Step-by-step explanation:

Given :-

• Slant height of frustum of cone = 8 cm
• Circumstances of circular end = 28 cm and 14 cm

To Find :-

CSA of frustum

Solution :-

We know that

Circumference of circle = 2πr

For circular end 1

2πr = 28

πr = 28/2

πr = 14

r = 14/π

For circular end 2

2πr' = 14

πr' = 14/2

πr' = 7

r' = 7/π

Now,

Total radius = R = r + r'

=> 14/π + 7/π

=> 14 + 7/π

=> 21/π

Now,

CSA of the frustum = πRl

=> 22/7 × (21/π) × 8

=> 22/7 × 21/(22/7) × 8

=> 22/7 × 21 × 7/22 × 8

=> 21 × 8

=> 168 cm²