Question

16A) The perpendiculae
sides of right triangle 6cm
8cm. if it is rotated
about its hypotoniese
then find the volume of
the double cone to formed​

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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