A right-angled triangle whose sides are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take π≃3.14]

A
3878 $c m^{3}$ , 1315.8 $c m^{2}$
B
3777 $c m^{3}$ , 1312.8 $c m^{2}$
C
3788 $c m^{3}$ , 1310.8 $c m^{2}$
D
3768 $c m^{3}$ , 1318.8 $c m^{2}$

Solution:

Given that,
Right-angled triangle with sides are 15 cm and 20 cm.

Let ABC be the right-angled triangle such that AB=15cm and AC=20cm
Using Pythagoras theorem in ΔABC we have,

\Rightarrow B C^{2} = A B^{2} + A C^{2}

\Rightarrow BC = \sqrt{(15)^{2} + (20)^{2}}

\Rightarrow BC = \sqrt{225}

\Rightarrow BC = 25 cm

Let OB=x and OA=y
Applying pythagoras theorem in triangles OAB and OAC we have,

A B^{2} = O B^{2} + O A^{2}; A C^{2} = O A^{2} + O C^{2}

\Rightarrow 1 5^{2} = x^{2} + y^{2}; 2 0^{2} = y^{2} + (25 - x)^{2}

\Rightarrow x^{2} + y^{2} = 225; (25 - x)^{2} + y^{2} = 400

{(25 - x)^{2} + y^{2}} - {x^{2} + y^{2}} = 400 - 225

\Rightarrow (25 - x)^{2} - x^{2} = 175

\Rightarrow x = 9

Putting x=9 in

x^{2} + y^{2} = 225

we get,

81 + y^{2} = 225

\Rightarrow y^{2} = 144

\Rightarrow y = 12

Thus we have OA=12cm and OB=9cm.
Volume of cone =

\frac{1}{3} π r^{2} h

Here, r is the radius and h is the height of cone.
Volume of the double cone = volume of cone CAA' + volume of cone BAA'
Volume of the double cone

= \frac{1}{3} π \times A O^{2} \times BO + \frac{1}{3} π \times A O^{2} \times CO

\Rightarrow \frac{1}{3} π \times A O^{2} (BO + CO)

\Rightarrow \frac{1}{3} π \times A O^{2} \times BC

\Rightarrow \frac{1}{3} \times 3.14 \times 12 \times 12 \times 25

\Rightarrow {3768 cm}^{3}

Curved surface area =

π \times r \times l

Here, l is slamt height.
Surface area of the double cone = curved surface area of cone CAA′
+ curved surface area of cone BAA’
Surface area of the double cone

= π \times AO \times AB + π \times AO \times AC

\Rightarrow π \times AO (AB + AC)

\Rightarrow 3.14 \times 12 \times (15 + 20)

\Rightarrow 3.14 \times 12 \times 35

\Rightarrow {1318.8 cm}^{2}

Therefore, the volume is

{3768 cm}^{3}

and the surface area of the double cone is

{1318.8 cm}^{2}

.
Hence, the correct option is 4.

A right-angled triangle whose sides are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take $π ≃ 3.14]$

Solution:

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