RMO 2006
How to Crack NTSE Stage-I Exam?
Every war starts with knowing what you are best at and where you lack. Start by doing a little SWAT analysis. Write down your weaknesses and strengths. Find out which subjects help you and which do not. After this now you know what are your loopholes that need a lot of work. And now work over … Continue reading How to Crack NTSE Stage-I Exam?
How to ace Kishore Vaigyanik Protsahan Yojana (KVPY)?
To score well in the Kishore Vaigyanik Protsahan Yojana, an on-going National Program of Fellowship in Basic Sciences by the Department of Science and Technology, one needs to follow a few pointers. To promote and encourage students with research-oriented careers in basic science courses, The Department of Science and Technology conducts a national level scholarship … Continue reading How to ace Kishore Vaigyanik Protsahan Yojana (KVPY)?
NTSE 2019-2020 Information, Dates, Pattern…
NTSE (National Talent Search Examination) 2019 is a National Level Examination conducted by NCERT (National Council of Education Research and Training) to provide the scholarship to the deserving candidates. This exam is for secondary school level to identify and recognize students with high intellect and skills. The examination is conducted every year at 2 levels: Stage … Continue reading NTSE 2019-2020 Information, Dates, Pattern…
JEE Main 2020: Eligibility, Schedule, Syllabus, Paper Pattern, Mock Tests and Important Dates
The Joint Entrance Examination (JEE) Mains is the national level undergraduate engineering entrance exam conducted to offer admissions in the undergraduate engineering programs at the National Institutes of Technology (NITs), the Indian Institutes of Information Technology (IIITs) and other centrally funded technical institutions (CFTIs) across the country. This exam is also the eligibility test for JEE Advanced which is … Continue reading JEE Main 2020: Eligibility, Schedule, Syllabus, Paper Pattern, Mock Tests and Important Dates
In a parallelogram ABCD the bisector of ∟A also bisects BC at P then how do you prove that AD = 2 AB?
Given AP is the angle bisector of ∠A and also bisects BC, so let ∠PAB=∠PAE=x and PB=PC=y Draw PE||BA =>∠PAB=∠APE=x {interior alternate angles} so, in ΔPAE, ∠PAE=∠APE=x according to the theorem that sides opposite to equal angles in a triangle are equal. =>AE=PE=y since, PE=AB=CD=y =>AD=2AB Hence Proved
