Question

If the radii of the circular ends of a frustum height 6cm are 15cm and 7cm respectively. Find the volume and curved surface area of a frustum.


gave meh correct answer with figure..​

Answer 1

Answer:

Given: Radii (r ) = 14 cm, and (r ) = 6 cm, height (h) = 6 cm Slant height of frustum (l) i. Curved surface area of frustum = πl (r1 + r2 ) = 3.14 × 10(14 + 6) = 3.14 × 10 × 20 = 628 cm2 ∴ The curved surface area of the frustum is 628 cm2 . ii. Total surface area of frustum = πl (r1 + r2 ) + πr12 + πr22 = 628 + 3.14 × (14)2 + 3.14 × (6)2 = 628 + 3.14 × 196 + 3.14 × 36 = 628 + 3.14(196 + 36) = 628 + 3.14 × 232 = 628 + 728.48 = 1356.48 cm2 ∴ The total surface area of the frustum is 1356.48 cm2. iii. Volume of frustum = 1/3 πth(r12 +r22 + r1 × r2 ) = 1/3 × 3.14 × 6(142 + 62 + 14 × 6) = 3.14 × 2(196 + 36 + 84) = 3.14 × 2 × 316 = 1984.48 cm3 ∴ The volume of the frustum is 1984.48 cm3.

Answer 2

Given :-

• Height of frustum, h = 6cm

• Radius of frustum of one end, r = 7 cm

• Radius of frustum of other end, R = 15 cm

To Find :-

• Volume of frustum

• Curved Surface Area of Frustum

Formula Used :-

$\rm :\longmapsto\:Volume_{(frustum)} = \dfrac{\pi \: h}{3}( {R}^{2} + {r}^{2} + Rr)$

$\rm :\longmapsto\:CSA_{(frustum)} = \pi \: (R + r) \: l$

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

where,

• R = radius of upper end of frustum

• r = radius of lower end of frustum

• h = height of frustum

• l = slant height is frustum

• CSA = Curved Surface Area

Solution :-

Given that,

• Height of frustum, h = 6cm

• Radius of frustum of one end, r = 7 cm

• Radius of frustum of other end, R = 15 cm

So,

Slant height of frustum is

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

$\rm :\longmapsto\:l \: = \: \sqrt{ {(15 - 7)}^{2} + {6}^{2} }$

$\rm :\longmapsto\:l \: = \: \sqrt{ {8}^{2} + {6}^{2} }$

$\rm :\longmapsto\:l \: = \: \sqrt{64 + 36}$

$\rm :\longmapsto\:l \: = \: \sqrt{100}$

$\rm :\longmapsto\:l \: = \: 10 \: cm$

Hence,

$\rm :\longmapsto\:CSA_{(frustum)} = \pi \: (R + r) \: l$

$\rm  \:  =  \: \:\dfrac{22}{7}(15 + 7) \times 10$

$\rm  \:  =  \: \:\dfrac{22}{7} \times 22 \times 10$

$\rm  \:  =  \: \:691.43 \: {cm}^{2}$

$\bf\implies \:CSA_{(frustum)}  =  \: \:691.43 \: {cm}^{2}$

Now,

We know,

$\rm :\longmapsto\:Volume_{(frustum)} = \dfrac{\pi \: h}{3}( {R}^{2} + {r}^{2} + Rr)$

$\rm  \:  =  \: \:\dfrac{1}{3} \times \dfrac{22}{7} \times 6 \times ( {15}^{2} + {7}^{2} + 15 \times 7)$

$\rm  \:  =  \: \: \dfrac{22}{7} \times 2 \times ( 225 + 49 + 105)$

$\rm  \:  =  \: \: \dfrac{44}{7} \times 379$

$\rm  \:  =  \: \:2382.28 \: {cm}^{3}$

$\bf\implies \:Volume_{(frustum)}  =  \: \:2382.28 \: {cm}^{3}$