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When a right-angled triangle is revolved around its hypotenuse, a double cone is formed with same radius but with different heights.

Given,

AB = 15 cm

AC = 20 cm

Let, OB = x and OA = y

Observe from the figure,

In right-angled triangle ABC, By Pythagoras theorem [Hypotenuse2 = Base2 + Perpendicular2]

BC2 = AC2 + AB2

BC2 = 202 + 152

BC2 = 400 + 225

BC2 = 625

BC = 25 cm

In ΔOAB

AB2 = OA2 + OB2

152 = x2 + y2 ……[1]

In ΔAOC

AC2 = OA2 + OC2

202 = y2 + (BC – OB)2

400 = y2 + (25 – x)2

400 = y2 + 625 – 50x + x2

400 = 152 + 625 – 50x

400 = 225 + 625 – 50x

50x = 450

x = 9 cm

Putting in [1], we get

152 = 92 + y2

y2 = 225 – 81

y2 = 144

y = 12 cm

Also, OC = 25 – x = 25 – 13 = 12 cm2

Now, Volume of cone

Where r denotes base radius and h denotes height of cone

Hence, volume of double cone

Curved surface area of cone = πrl

Where r denotes base radius and l denotes slant height

Also,

Surface area of double cone = CSA of left cone + CSA of right cone

Putting values,

Surface area = π(OA)(AB) + π(OA)(AC)

= 3.14 × 12 × (15 + 20)

= 1318.8 cm2

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