Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The vertices of a triangle are <img alt="\mathrm{(0,0),(a, 0) and (0, b).}" src="https://

The vertices of a triangle \mathrm{O A B} are \mathrm{(0,0),(a, 0) and (0, b).} Then, the distance between the orthocentre and circumcentre of the triangle \mathrm{O A B} is
 

Option: 1

\mathrm{a+b}

 


Option: 2

\mathrm{a-b}
 


Option: 3

\mathrm{\frac{1}{2} \sqrt{a^2+b^2}}
 


Option: 4

\mathrm{\frac{1}{2}(a+b)}


Answers (1)

best_answer

For a right-angled triangle \mathrm{A O B}, the orthocentre is \mathrm{O(0,0)} and circumcentre is the middle point of hypotenuse \mathrm{A B.}
\Rightarrow the circumcentre is \mathrm{P\left(\frac{a}{2}, \frac{b}{2}\right)}
Hence, the required distance \mathrm{O P=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}=\frac{1}{2} \sqrt{a^2+b^2}}

Hence option 3 is correct.
 

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 registration last day, city slip soon; here's how to prepare
anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

Latest Question