Q. 1

The radii of the ends of a frustum of a cone 45 cm high are 28cm and 7cm. Find its volume, the curved surface area. [Take π=22/7]

Answer :


Frustum = difference of two right circular cones OAB and OCD


Let the height of the cone OAB be h1 and its slant height l1


i.e. OA = OB = l1 and OP = h1.


Let h2 be the height of cone OCD and l2 its slant height


i.e. OC = OD = l2 and OQ = h2


We have , r1 = 28cm and r2 = 7cm


and height of frustum (h) = 45cm


Also,


h1 = 45 + h2 …(i)


Now, we first need to determine the h1 and h2


ΔOPB and OQD are similar, we have



h1 = 4h2 …(ii)


From (i) and (ii), we get


4h2 = 45 + h2


3h2 = 45


h2 = 15cm


Putting the value of h2 in eq. (ii), we get


h1 = 4×15 = 60cm


So, h1 = 60cm and h2 = 15cm


Now, the volume of frustum = Vol. of cone OAB – Vol. of cone OCD







= 5 × 22 × 441


= 48510 cm3


Now, we first have to find the slant height l1 and l2


l1 = √{(28)2+(60)2}


l1 = √{(4×7)2 + (4×15)2


l1 = 4√{(7)2 + (15)2}


l1 = 4√49 + 225


l1 = 4√(274)


l1 = 4 × 16.55


l1 = 66.20cm


and l2 = √{(7)2 + (15)2}


l2 = √49 + 225


l2 = √(274)


l2 = 16.55 cm


So, CSA of frustum = CSA of cone OAB – CSA of cone OCD


= πr1l1 – πr2l2



= 22 [4×66.20 – 16.55]


= 22 × 248.25


= 5461.5cm2


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