Question

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Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Correct \: Question}}} \mid}}$

The perpendicular sides of right triangle are 6cm and 8cm. if it is rotated about its hypotenuse then find the volume of the double cone to formed?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Let AB = 8 cm.

BC = 6 cm.

The triangle is right angled at B I.e. ∠B = 90°.

$\huge{\bold{\underline{Explanation:-}}}$

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Firstly We need to Find The Hypotenuse of The triangle.

Now,

From Pythagoras theorem,

$\large{\boxed{\tt AC^2 = AB^2 + BC^2}}$

Substituting the values,

$\large{\tt \leadsto (AC)^2 = (8)^2 + (6)^2}$

$\large{\tt \leadsto (AC)^2 = 64 + 36}$

$\large{\tt \leadsto (AC)^2 = 100}$

$\large{\tt \leadsto AC = \sqrt{100}}$

$\large{\boxed{\tt AC = 10 \: Cm}}$

So, The Hypotenuse of the Right Triangle is 10 cm.

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Finding The radius of the Double Cone.

Now, The radius of the Double Cone formed is r = OB .

We know,

$\large{\boxed{\tt Area \: of \: \Delta ABC = \dfrac{1}{2} \times AB \times BC}}$

But,

Area of ΔABC = ½ × BO × AC

(Viewing From The Double Cone formed)

Equating Both ,

$\large{\tt \leadsto \dfrac{1}{2} \times BO \times AC = \dfrac{1}{2} \times AB \times BC}$

Substituting the values,

$\large{\tt \leadsto \dfrac{1}{2} \times OB \times 10 = \dfrac{1}{2} \times 6 \times 8}$

$\large{\tt \leadsto \cancel{\dfrac{1}{2}} \times OB \times 10 = \cancel{\dfrac{1}{2}} \times 6 \times 8}$

$\large{\tt \leadsto OB \times 10 = 48}$

$\large{\tt \leadsto OB = \dfrac{48}{10}}$

$\large{\boxed{\tt OB = 4.8 cm}}$

So, the Radius (r) of the Double Cone is 4.8 cm.

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After Rolling The Double Cone is formed (Refer the attachment for figure)

Now, Finding the Volume of the Double Cone Formed.

$\large{\boxed{\tt Volume \: of \: Double \: cone = Volume \: of \: Cone_{(1)} + Volume \: of \: Cone_{(2)}}}$

Substituting the values,

$\large{\tt \leadsto V = \dfrac{1}{3} \pi r^2 h_{(1)}+ \dfrac{1}{3} \pi r^2 h_{(2)}}$

Here h₁ and h₂ are the Heights of Cone From the Center of the double Cone.

Now,

$\large{\tt \leadsto V = \dfrac{1}{3} \pi r^2 \bigg( h_1 + h_2 \bigg)}$

$\large{\tt \leadsto V = \dfrac{1}{3} \pi r^2 \bigg(OA + OC \bigg)}$

$\large{\tt \leadsto V = \dfrac{1}{3} \pi r^2 \bigg(10 \bigg)}$

As OA + OC = AC Which is equal to Hypotenuse of the Triangle.

$\large{\tt \leadsto V = \dfrac{1}{3} \times \dfrac{22}{7} \times (4.8)^2 \bigg(10 \bigg)}$

$\large{\tt \leadsto V = \dfrac{1}{\cancel{3}} \times \dfrac{22}{7} \times \cancel{23.04} \times (10)}$

$\large{\tt \leadsto V = \dfrac{22}{7} \times 7.68 \times 10}$

$\large{\tt \leadsto V = \dfrac{22}{7} \times 76.8}$

Simplifying,

$\huge{\boxed{\boxed{\tt Volume = 241.37 \: cm^2}}}$

So, the Volume of the Double Cone formed is 241.37 cm².

#refer the attachment for figure.

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Answer 2

Answer:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\sf{Question}$

$\small \sf The \: perpendicular \: sides \: of \: right \: triangle \: \\ \sf are \: 6cm \: and \: 8cm \: . \\ \small \sf \: if \: it \: is \: rotated \: about \: its \: hypotenuse \: then \: \\ \small \sf \: find \: the \: volume \: of \: the \: double \: cone \: to \:formed \: ?$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bcancel{\large{\blue{\fbox{\tt{αɴѕωєя࿐}}}}}$

The double cone so formed the right angled triangle ABC 

Hypotenuse AC = underoot 6 square + 8 square  

❥ underoot 36 + 64 

❥ underoot 100 

❥ 10 cm  

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Area of ΔABC = ½ x AB x AC 

½ x AC x OB = ½ x 6 x 8 

½ x 10 x OB = 24 

OB = 4.8 cm  

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Volume of double cone = volume of cone 1 + volume  of cone 2  

❥ 1/3 π r square h1 + 1/3π r square h2 

❥ 1/3π r square (h1+h2) 

❥ 1/3π r square (OA+OC) 

❥ 1/3 x 22/7 x (4.8) square x 10 

❥ 241.3 cm cube 

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Surface area of double cone = surface area of cone 1 + surface area of cone 2 

π rl1 + π rl2 

π r (6+8) 

❥ 22/7 x 4.8 x 14 

❥ 211.2 cm square

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