Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let a and b represent the length of a right triangle's legs. If d is the diameter of a circle inscribed into the triangle, and D is the diameter of a circle superscribed on the triangle, then d

Let a and b represent the length of a right triangle's legs. If d is the diameter of a circle inscribed into the triangle, and D is the diameter of a circle superscribed on the triangle, then d + D equals

Option: 1

\mathrm{a}+\mathrm{b}


Option: 2

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


Option: 3

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


Option: 4

\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}


Answers (1)

best_answer

\mathrm{AB}=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}

\mathrm{hence \, D=\sqrt{b^{2}+a^{2}}}\quad \cdots(1)

\mathrm{Now \, \frac{\mathrm{d}}{2}=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{ab}}{2 \mathrm{s}}}
\mathrm{\therefore \quad \frac{\mathrm{d}}{2}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}} \text { or } \quad \mathrm{d}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}}\quad \cdots(2)

from (1) and (2)

\mathrm{d+D=\frac{\sqrt{a^{2}+b^{2}}\left[(a+b)+\sqrt{a^{2}+b^{2}}\right]+2 a b}{a+b+\sqrt{a^{2}+b^{2}}}=\frac{(a+b)^{2}+(a+b) \sqrt{a^{2}+b^{2}}}{a+b+\sqrt{a^{2}+b^{2}}}} \Rightarrow \mathrm{(A)}

Posted by

Divya Prakash Singh

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 city slip soon; admit card, exam dates, 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