Math, asked by Heeranh, 11 months ago

5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two
parts at the middle of its height by a plane parallel to its base. If the frustum so obtained
be drawn into a wire of diameter - cm, find the length of the wire.
16.

Answers

Answered by Anonymous

Answer:

7964.4 m

Step-by-step explanation:

Let R be the radius of the base of the cone and r be the radius of the upper of frustum.

Then,

In triangle ABG

$\tan(30) = \frac{R \: }{AG \: }$

$\frac{1}{ \sqrt{3} } = \frac{R \: }{20} \\ \\ R = \frac{20}{ \sqrt{3} } \\ \\ R = \frac{20 \sqrt{3} }{3}$

______

In triangle ADF

$\tan(30) = \frac{r \: }{AF \: } \\ \\ \frac{1}{ \sqrt{3} } = \frac{r}{10} \\ \\ r = \frac{10}{ \sqrt{3} } \\ \\ r = \frac{10 \sqrt{3} }{3} \:$

________

Height of the frustum, h =10cm (given)

Volume of the used in frustum

$= \frac{1}{3} \pi \: h( {R}^{2} + {r}^{2} +R r)$

$= \frac{1}{3} \pi \times 10( {( \frac{20 \sqrt{3} }{3} \: )}^{2} + ( \frac{ {10 \sqrt{3} }^{2} }{3} ) + \frac{20 \sqrt{3} \times 10 \sqrt{3} }{3} ) \\ \\ = \frac{1}{3} \pi \times 10( \frac{400}{3} + \frac{100}{3} + \frac{200}{3} ) \\ \\ = \frac{7000}{9} \pi \: {cm}^{3} \\ \\$

_______

Now the radius of wire = 1/32 cm

Let l be the length of the wire.

Then the volume of the cylindrical wire = volume of metal used in frustum.

$\pi { \frac{1}{32} }^{2} \times l = \frac{7000}{9} \pi \\ \\ l = \frac{7000 \times 32 \times 32}{9} \\ \\ l = \frac{7168000}{9} cm \\ \\ l = \frac{7168000}{9 \times 100} \\ \\ = 7964.4m$

Therefore : length of the wire is 7964.4m

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7964.4 m

Let R be the radius of the base of the cone and r be the radius of the upper of frustum.

Then,

In triangle ABG

\tan(30) = \frac{R \: }{AG \: }tan(30)=AGR

$$\begin{lgathered}\frac{1}{ \sqrt{3} } = \frac{R \: }{20} \\ \\ R = \frac{20}{ \sqrt{3} } \\ \\ R = \frac{20 \sqrt{3} }{3}\end{lgathered}$$

______

In triangle ADF

$$\begin{lgathered}\tan(30) = \frac{r \: }{AF \: } \\ \\ \frac{1}{ \sqrt{3} } = \frac{r}{10} \\ \\ r = \frac{10}{ \sqrt{3} } \\ \\ r = \frac{10 \sqrt{3} }{3} \:\end{lgathered}$$

________

Height of the frustum, h =10cm (given)

Volume of the used in frustum

$$= \frac{1}{3} \pi \: h( {R}^{2} + {r}^{2} +R r)$$

$$\begin{lgathered}= \frac{1}{3} \pi \times 10( {( \frac{20 \sqrt{3} }{3} \: )}^{2} + ( \frac{ {10 \sqrt{3} }^{2} }{3} ) + \frac{20 \sqrt{3} \times 10 \sqrt{3} }{3} ) \\ \\ = \frac{1}{3} \pi \times 10( \frac{400}{3} + \frac{100}{3} + \frac{200}{3} ) \\ \\ = \frac{7000}{9} \pi \: {cm}^{3} \\ \\\end{lgathered}$$

_______

Now the radius of wire = 1/32 cm

Let l be the length of the wire.

Then the volume of the cylindrical wire = volume of metal used in frustum.

$$\begin{lgathered}\pi { \frac{1}{32} }^{2} \times l = \frac{7000}{9} \pi \\ \\ l = \frac{7000 \times 32 \times 32}{9} \\ \\ l = \frac{7168000}{9} cm \\ \\ l = \frac{7168000}{9 \times 100} \\ \\ = 7964.4m\end{lgathered}$$

Therefore : length of the wire is 7964.4m

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5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into twoparts at the middle of its height by a plane parallel to its base. If the frustum so obtainedbe drawn into a wire of diameter - cm, find the length of the wire.16. ​

Answers

5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two
parts at the middle of its height by a plane parallel to its base. If the frustum so obtained
be drawn into a wire of diameter - cm, find the length of the wire.
16.