- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
In figure below, D is the mid-point of side BC and $AE \perp BC$. If $ B C=a, A C=b, A B=C, E D=x, A D=p $ and $ A E=h, $ prove that $ c^{2}=p^{2}-a x+\frac{a^{2}}{4} $.
"
Given:
In the given figure, D is the mid-point of side BC and $AE \perp BC$.
\( B C=a, A C=b, A B=C, E D=x, A D=p \) and \( A E=h \).
To do:
We have to prove that \( c^{2}=p^{2}-a x+\frac{a^{2}}{4} \).
Solution:
In $\triangle AED$, by using Pythagoras theorem,
$AD^2=AE^2+ED^2$
$AE^2=AD^2-ED^2$.....(i)
In $\triangle AEB$, by using Pythagoras theorem,
$AB^2=AE^2+BE^2$
$c^2=(AD^2-ED^2)+(BD-ED)^2$ (From (i) and $BE=BD-ED$)
$c^2=AD^2-ED^2+BD^2+ED^2-2BD\times ED$
$c^2=AD^2+BD^2-2BD\times ED$
$c^2=p^2+(\frac{a}{2})^2-2\times(\frac{a}{2})\times x$ (Since $DC=\frac{BC}{2}$)
$c^2=p^2+\frac{a^2}{4}-ax$
Hence proved.
Advertisements