Question

A double cone is formed by revolving a right angled triangle having sides 5cm, 12cm and 13cm about its hypotenuse. Find total surface area and volume of double cone. Answer this question and get 50 points!!!​

Answer 1

Total surface area = 246.5 cm² & Total Volume = 290 cm³ if Double cone is formed by revolving right angled triangle having sides 5cm, 12cm and 13cm about its hypotenuse

Step-by-step explanation:

Right angle triangle is revolved about the hypotenuse AC, the figure so formed is a double cone.

Attached is figure

In triangle ABC,

AB = 5 cm

AC = 12 cm

BC = 13 cm

OB = h₁

OC = h₂

OA = OD = r

h₁ + h₂  = 13

5² =h₁²  + r²

12² = h₂² + r²

=> 12² - 5² = h₂² - h₁²

=> 144 - 25 = (h₂+ h₁)(h₂ - h₁)

=> 119 = 13(h₂ - h₁)

=> h₂ - h₁ =119/13

h₁ + h₂  = 13

=> h₁ = 25/13

=> h₂ = 144/13

5² =h₁²  + r²  or 12² = h₂² + r²

=> r² = 25 - (625/169)   or 144 - (144²/169)

=> r² = 25 * 144/169

=> r = 5 * 12 /13

=> r = 60/13

total surface area = Curved surface are of both

= π * (60/13) * 5  + π * (60/13) * 12

= π * 60 * 17 /13

= 246.5 cm²

Total Volume = Volume of both cone

= (1/3)π (60/13)² * (25/13)  + (1/3)π (60/13)² * (144/13)

= ( (1/3)π 3600 / 169 )  * 13

= 1200 π / 13

= 290 cm³

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Answer 2

Hello Dear User,

Question :-

A double cone is formed by revolving a right angled triangle having sides 5cm, 12cm and 13cm about its hypotenuse. Find total surface area and volume of double cone.

Answer :-

Total surface area = 246.5 cm²

Total Volume = 290 cm³

Explanation :-

Given :

$\setlength{\unitlength}{.4in} \begin{picture}(7,5)(0,0) \linethickness{1pt} \put(0,0){\line(1,0){4}} \put(4,0){\line(0,1){3}} \put(0,0){\line(4,3){4}} \put(2,-.25){\makebox(0,0){BC}} \put(4.25,1.5){\makebox(0,0){AB}} \put(2,2){\makebox(0,0){AC}} \end{picture}$

AB = 5 cm

AC = 12 cm            

BC = 13 cm

Let :

OB = h₁

OC = h₂

-----------------------------------

OA = OD = r

h₁ + h₂  = 13  (A.T.Q)

5² =h₁²  + r²

12² = h₂² + r²

=> 12² - 5² = h₂² - h₁²

=> 144 - 25 = (h₂+ h₁)(h₂ - h₁)

=> 119 = 13(h₂ - h₁)

=> h₂ - h₁ =119/13

h₁ + h₂  = 13

=> h₁ = 25/13

=> h₂ = 144/13

5² =h₁²  + r²  or 12² = h₂² + r²

=> r² = 25 - (625/169)   or 144 - (144²/169)

=> r² = 25 * 144/169

=> r = 5 * 12 /13

=> r = 60/13

------------------------------------------------

total surface area = Curved surface are of both

= π * (60/13) * 5  + π * (60/13) * 12

= π * 60 * 17 /13

= 246.5 cm²

------------------------------------------------

Total Volume = Volume of both cone

= (1/3)π (60/13)² * (25/13)  + (1/3)π (60/13)² * (144/13)

= ( (1/3)π 3600 / 169 )  * 13

= 1200 π / 13

= 290 cm³

Additional Information :-

Glad to help