Answer :

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 [Hypotenuse^{2} = Base^{2} + Perpendicular^{2}]

BC^{2} = AC^{2} + AB^{2}

⇒ BC^{2} = 20^{2} + 15^{2}

⇒ BC^{2} = 400 + 225

⇒ BC^{2} = 625

⇒ BC = 25 cm

In ΔOAB

AB^{2} = OA^{2} + OB^{2}

⇒ 15^{2} = x^{2} + y^{2} ……[1]

In ΔAOC

AC^{2} = OA^{2} + OC^{2}

⇒ 20^{2} = y^{2} + (BC – OB)^{2}

⇒ 400 = y^{2} + (25 – x)^{2}

⇒ 400 = y^{2} + 625 – 50x + x^{2}

⇒ 400 = 15^{2} + 625 – 50x

⇒ 400 = 225 + 625 – 50x

⇒ 50x = 450

⇒ x = 9 cm

Putting in [1], we get

15^{2} = 9^{2} + y^{2}

⇒ y^{2} = 225 – 81

⇒ y^{2} = 144

⇒ y = 12 cm

Also, OC = 25 – x = 25 – 13 = 12 cm^{2}

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 cm^{2}

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