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]

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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