Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Palak Yadav 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
{tex}\Delta A B C{/tex} and {tex}\triangle A D E{/tex} are similar triangles and hence the corresponding sides are proportional.
B is the mid-point of AD.
Thus, {tex}\frac { A B } { A D } = \frac { B C } { D E }{/tex}
Let r and R be the radii of both the ends of the frustum.
So, {tex}\frac { 10 } { 20 } = \frac { r } { R }{/tex}
{tex}\Rightarrow{/tex} R = 2r
Also consider the {tex}\Delta A D E{/tex}
{tex}\tan 30 ^ { \circ } = \frac { \text { opposite side } } { \text { adjacent side } }{/tex}
{tex}\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { R } { 20 }{/tex}
{tex}\Rightarrow R = \frac { 20 } { \sqrt { 3 } } \mathrm { cm }{/tex}
{tex}\Rightarrow r = \frac { 10 } { \sqrt { 3 } } \mathrm { cm }{/tex}
Volume of the frustum {tex}= \frac { \pi } { 3 } \left[ R ^ { 2 } + R r + r ^ { 2 } \right] h{/tex}
{tex}\Rightarrow{/tex} {tex}\pi \left( \frac { \frac { 1 } { 12 } } { 2 } \right) ^ { 2 } h = \frac { \pi } { 3 } \left[ \frac { 400 } { 3 } + \frac { 200 } { 3 } + \frac { 100 } { 3 } \right] \times 10{/tex}
{tex}\Rightarrow \frac { 1 } { 576 } \times h = \frac { 1 } { 3 } \left[ \frac { 700 } { 3 } \right] \times 10{/tex}
{tex}\Rightarrow{/tex} {tex}h = \frac { 1 } { 3 } \left[ \frac { 700 } { 3 } \right] \times 10 \times 576{/tex}
{tex}\Rightarrow{/tex} h = 100{tex}\times{/tex}10{tex}\times{/tex}576
{tex}\Rightarrow{/tex} h= 448000 cm
{tex}\Rightarrow{/tex} h= 4480 m
Hence, the length of the wire is 4480 m.
Posted by Mohsin Shaikh 6 years, 11 months ago
- 1 answers
Posted by Asha Singh 6 years, 11 months ago
- 1 answers
Posted by Mihir Arya 6 years, 11 months ago
- 1 answers
Posted by Manish Bairwa 6 years, 11 months ago
- 1 answers
Posted by Yash Patil 6 years, 11 months ago
- 1 answers
Chētnà Pandey✌️ 6 years, 11 months ago
Posted by Shubham Mishra 6 years, 11 months ago
- 1 answers
Posted by Shanaya Gurjar 6 years, 11 months ago
- 0 answers
Posted by Suchita Meena 6 years, 11 months ago
- 1 answers
Simar Simar 6 years, 11 months ago
Posted by @Kumu...#Vk... ?? 6 years, 11 months ago
- 3 answers
Nidhi Chhonker 6 years, 11 months ago
Simar Simar 6 years, 11 months ago
Shiva Rawat 6 years, 11 months ago
Posted by Sameer Ali 6 years, 11 months ago
- 1 answers
?☠Debraj Bairagi☠? 6 years, 11 months ago
Posted by Sumit Jangra 6 years, 11 months ago
- 4 answers
Sonal Kumari 6 years, 11 months ago
Poonam Bais 6 years, 11 months ago
Posted by Sameer Ali 6 years, 11 months ago
- 2 answers
Posted by S Sihag Ji 6 years, 11 months ago
- 7 answers
Saumya Saini 6 years, 11 months ago
Posted by Abhishek Kandari 6 years, 11 months ago
- 3 answers
Posted by Abhishek Kandari 6 years, 11 months ago
- 3 answers
Sushma Bhattoo 6 years, 11 months ago
Posted by Abhinav Chandel 6 years, 11 months ago
- 0 answers
Posted by Puja Sahoo 6 years, 11 months ago
- 0 answers
Posted by Nadeem Akhtar 6 years, 11 months ago
- 0 answers
Posted by Puja Sahoo 6 years, 11 months ago
- 0 answers
Posted by ???????? ???? 6 years, 11 months ago
- 5 answers
Karuna Sharma 6 years, 11 months ago
Posted by Thakur Subham Singh Chauhan 6 years, 11 months ago
- 0 answers
Posted by Pulkit Singh 6 years, 11 months ago
- 2 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 6 months ago
Let ABC be the equilateral triangle such that,
A = (3,0), B=(6,0) and C=(x,y)
Distance between:
{tex}\sqrt {( x_{2}-x_{1})^2+(y_{2} -y_{1})^{2} }{/tex}
we know that,
AB=BC=AC
By distance formula we get,
AB=BC=AC=3units
AC=BC
{tex}\sqrt{(3-x)^2+y^2}=\sqrt{(6-x)^2+y^2}{/tex}
{tex}9+x^2-6 x+y^2=36+x^2-12 x+y^2{/tex}
{tex}6 x=27{/tex}
{tex}x=27 / 6=9 / 2{/tex}
BC = 3 units
{tex}\sqrt{(6-\frac{27}{6})^2+y^2}=3{/tex}
{tex}(\frac{(36-27)}{6})^2+y^2=9{/tex}
{tex}(\frac{9}{6})^2+y^2=9{/tex}
{tex}(\frac{3}{2})^2+y^2=9{/tex}
{tex}\frac{9}{4}+y^2=9{/tex}
{tex}9+4 y^2=36{/tex}
{tex}4 y^2=27{/tex}
{tex}y^2=\frac{27}{4}{/tex}
{tex}y=\sqrt{(\frac{27}{4})}{/tex}
{tex}y=3 \sqrt{\frac{3}{2}}{/tex}
{tex}(x, y)=(9 / 2,3 \sqrt{\frac{3}{2}}){/tex}
Hence third vertex of equilateral triangle = C = {tex}(9 / 2,3 \sqrt{\frac{3}{2}}){/tex}
0Thank You