Question

The radio of the ends of a frustum of a cone 45cm high are 28 CM and 7cm .
Did it's volumes,curved surface area and total surface area.

Answer 1

Answer:

Given,  

Height of the bucket (h) = 45 cm

Radius one (R) = 28 cm

Radius a pair of (r) = 7 cm

Volume of solid = π/3 h (R²+r²+R×r)

where,  

R=  larger radius  

r = smaller radius  

h=  height

=> π/3 forty five [(28)² + (7)² + 28×7]

=> π × 15× [ 784 + 49 + 196 ]

=> π × 15 × 1029

=> π × 15435

=> 22/7 × 15435

=> 22× 2205

= > 48510 cm³

Hence our volume is 48510 cm³

We have

=> l =$\sqrt{h^{2} +(r_{1} - r_{2})2 } = \sqrt{(45^{2} )+(28-7)^{2} } cm$

=> 3$\sqrt{(15)^{2} + (7)^{2}  } =49.65 cm$

so the curved surface area of the frustum

=> $\pi (r_{1} +r_{2}) l = 22/7 (28+7) (49.65) \\ =5461.5 cm^{2}$

=> Total curved area surface of the frustum

= $\pi (r_{1} +r_{2} )l+ \pi r^{2} _{1} +  \pi r^{2} _{2}$

=[5461.5+22/7 × (28)$^{2}$] +22/7 × (7)$^{2}$

=>8079.5 cm$^{2}$

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