Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 188048 by Rupesh123 last updated on 25/Feb/23

Answered by mr W last updated on 25/Feb/23

say A=area, r=inradius A=((ab)/2) c=(√(a^2 +b^2 )) r=((2A)/(a+b+c))=((ab)/(a+b+(√(a^2 +b^2 )))) ≤((ab)/(2(√(ab))+(√(2ab))))=((√(ab))/( (√2)((√2)+1))) r^2 ≤((ab)/(2((√2)+1)^2 )) =(A/(((√2)+1)^2 ))=(A/(3+2(√2)))<(A/(3+2))=(A/5) ⇒A>5r^2

sayA=area,r=inradiusA=ab2c=a2+b2r=2Aa+b+c=aba+b+a2+b2⩽ab2ab+2ab=ab2(2+1)r2⩽ab2(2+1)2=A(2+1)2=A3+22<A3+2=A5⇒A>5r2

Commented by Rupesh123 last updated on 25/Feb/23

Great!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com