Question

AP perpendicular to QR AND PR>PQ SHOW THAT AR>AQ

Answer 1

from pythagoras theorem
${qp}^{2} = {qa}^{2} + {ap}^{2}$
again in triangle PAR
${pr}^{2} = {ar}^{2} + {ap}^{2}$
cancel ap square from both equation
${pr}^{2} - {qp}^{2} = {ar}^{2} - {qa}^{2}$
given pr>qp
then ar>qa from the above equation

Answer 2

Thank you for asking this question. Here is your answer:

PR > PQ

Now we will apply Pythagoras theorem in triangle APQ,

AP² = AQ² - PQ²      this will be the equation 1

now we will Apply Pythagoras theorem in triangle APR,

AP² = AR² - PR²           and this will be the equation 2

AR² - PR² = AQ² - PQ²

Given PR > PQ

AR² - PQ² > AQ² - PQ²

AR²  >  AQ²

AR > AQ

If there is any confusion please leave a comment below.