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Question Number 53962 by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

Find maximum perimeter of   triangle APQ.(have i posted this  question, before, i′m not sure!?)

$${Find}\:{maximum}\:{perimeter}\:{of}\: \\ $$$${triangle}\:{APQ}.\left({have}\:{i}\:{posted}\:{this}\right. \\ $$$$\left.{question},\:{before},\:{i}'{m}\:{not}\:{sure}!?\right) \\ $$

Commented by ajfour last updated on 27/Jan/19

is it just = a+b+(√(a^2 +b^2 ))  ?

$${is}\:{it}\:{just}\:=\:{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:? \\ $$

Commented by mr W last updated on 27/Jan/19

yes sir.

$${yes}\:{sir}. \\ $$

Answered by mr W last updated on 27/Jan/19

Commented by mr W last updated on 27/Jan/19

A′D=b  A′′B=a  AQ=A′Q  PA=PA′′  perimeter of ΔAPQ=A′Q+QP+PA′′    min. perimeter=A′A′′=2(√(a^2 +b^2 ))    max. perimeter=A′D+DB+BA′′  =b+(√(a^2 +b^2 ))+a

$${A}'{D}={b} \\ $$$${A}''{B}={a} \\ $$$${AQ}={A}'{Q} \\ $$$${PA}={PA}'' \\ $$$${perimeter}\:{of}\:\Delta{APQ}={A}'{Q}+{QP}+{PA}'' \\ $$$$ \\ $$$${min}.\:{perimeter}={A}'{A}''=\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$${max}.\:{perimeter}={A}'{D}+{DB}+{BA}'' \\ $$$$={b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{a} \\ $$

Commented by ajfour last updated on 28/Jan/19

great technique to prove it Sir, thank you!

$${great}\:{technique}\:{to}\:{prove}\:{it}\:{Sir},\:{thank}\:{you}! \\ $$

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