Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Arjav Jain 3 months, 3 weeks ago
- 0 answers
Posted by Akul Singh 11 months, 1 week ago
- 0 answers
Posted by Sheral Jain 11 months, 4 weeks ago
- 0 answers
Posted by R. S Jokers 1 year, 1 month ago
- 0 answers
Posted by Pragya Kala 1 year, 2 months ago
- 2 answers
R. S Jokers 1 year, 1 month ago
Posted by Parbhuram. Bamniya 1 year, 3 months ago
- 1 answers
Posted by Shreya Verma 1 year, 3 months ago
- 2 answers
Posted by Shreya Verma 1 year, 3 months ago
- 1 answers
Posted by Naveen Gora 1 year, 5 months ago
- 1 answers
R. S Jokers 1 year, 2 months ago
Posted by Anmol Rajak 1 year, 5 months ago
- 2 answers
R. S Jokers 1 year, 1 month ago
Few rules to keep homework help section safe, clean and informative.
- Don't post personal information, mobile numbers and other details.
- Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
- Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
- Ask specific question which are clear and concise.
Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.
Posted by Kittu Jaat 1 year, 7 months ago
- 4 answers
Posted by Piyush Kumar 1 year, 8 months ago
- 1 answers
Preeti Dabral 1 year, 8 months ago
Given: {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}DEF
AM is a median in {tex}\triangle {/tex}ABC and DN is the corresponding median in {tex}\triangle {/tex}DEF
To prove: {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}
Proof: {tex}\triangle {/tex}DBC {tex} \sim {/tex} {tex}\triangle {/tex}DEF
{tex}\Rightarrow {/tex} {tex}\angle{/tex} A = {tex}\angle{/tex} D, {tex}\angle{/tex} B = {tex}\angle{/tex} E, {tex}\angle{/tex} C = {tex}\angle{/tex} F and
{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{AC}}{{DF}}{/tex}
Also, {tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}}{/tex}={tex}\frac{{A{B^2}}}{{D{E^2}}} = \frac{{B{C^2}}}{{E{F^2}}} = \frac{{A{C^2}}}{{D{F^2}}}{/tex}........(i)[area theorem]
Now,{tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} = \frac{{\frac{1}{2}BC}}{{\frac{1}{2}EF}} = \frac{{BM}}{{EN}}{/tex}..............(ii)
In {tex}\triangle {/tex}ABM and DEN
{tex}\angle{/tex} B= {tex}\angle{/tex} E and {tex}\frac{{AB}}{{DE}} = \frac{{BM}}{{EN}}{/tex} [From (ii)]
{tex}\Rightarrow {/tex} {tex}\triangle {/tex}ABM {tex} \sim {/tex}{tex}\triangle {/tex}DEN [SAS similarity]
{tex}\Rightarrow {/tex}{tex}\frac{{area\vartriangle ABM}}{{area\vartriangle DEN}} = \frac{{A{B^2}}}{{D{E^2}}} = \frac{{A{M^2}}}{{D{N^2}}} = \frac{{B{M^2}}}{{E{N^2}}}{/tex}............(iii)
From (i) and (iii), we get
{tex}\frac{{area\vartriangle ABC}}{{area\vartriangle DEF}} = \frac{{A{M^2}}}{{D{N^2}}}{/tex}
Posted by Ramfool Saini 1 year, 9 months ago
- 0 answers
Posted by Devkaran Jaat 1 year, 9 months ago
- 5 answers
Posted by Rinku Parihar 1 year, 9 months ago
- 4 answers
Posted by Uv R 1 year, 10 months ago
- 2 answers
Posted by Rakesh Nayak 1 year, 10 months ago
- 1 answers
Posted by Gaming With Happy 1 year, 10 months ago
- 2 answers
Posted by Lavish Chhabra 1 year, 11 months ago
- 0 answers
Posted by Kashish Jain 2 years ago
- 0 answers
Posted by Neha Saran 2 years ago
- 0 answers
Posted by Heena Kosar 2 years, 2 months ago
- 1 answers
Posted by Antika Gaur 2 years, 2 months ago
- 1 answers
Posted by Rudra Rao 2 years, 2 months ago
- 4 answers
Posted by Komal Godara 2 years, 2 months ago
- 0 answers
Posted by Vinay Pratap 2 years, 4 months ago
- 1 answers
Posted by Abhishek Nayak 2 years, 5 months ago
- 0 answers
Posted by Shubham Singh 1 year, 8 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
