Question

the radius of frustum of a cone 45cm high are 28 cm and 7 cm . find its volume , curved surface area and total surface area

Answer 1

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Answer 2

Answer:

The volume of the frustum , curved surface area and total surface area is  $48510 cm^3$ ,  $5462.468 cm^2$ and $8080.468 cm^2$ respectively.

Step-by-step explanation:

Radius of upper circular base r = 7 cm

Radius of lower circular base R = 28 cm

Height of frustum = 45 cm

Volume of frustum = $\frac{1}{3} \times \pi \times h(r^2 + R^2 + r \times R)$

                               = $\frac{1}{3} \times \frac{22}{7} \times 45(7^2 + 28^2 + 7 \times 28)$

                               = $48510 cm^3$

Slant height = $l=\sqrt{h^2+(R-r)^2}$

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

                   = $l=49.6588$

Curved surface area of frustum = $\pi \times  l(R + r)$

                                                    = $\frac{22}{7} \times 49.6588(28+7)$

                                                    = $5462.468 cm^2$

Total Surface area of frustum = $\pi \times l(R + r) + \pi(R^2 + r^2)$

                                                 = $\frac{22}{7} \times 49.6588(28+7) + \frac{22}{7}(28^2 + 7^2)$

                                                 = $8080.468 cm^2$

Hence the volume of the frustum , curved surface area and total surface area is  $48510 cm^3$ ,  $5462.468 cm^2$ and $8080.468 cm^2$ respectively.