Question 1011929
In quadrilateral {{{PQRS}}} ,
vertex {{{P}}} is adjacent to vertices {{{Q}}} and {{{S}}} .
Since there is a right angle at {{{P}}} ,
triangle {{{PQS}}} is a right triangle with hypotenuse {{{QS}}} ,
and legs {{{PQ=9}}} and {{{PS=12}}} .
Using the Pytahagorean theorem, we can find hypotenuse {{{QS}}} :
{{{QS^2=PQ^2+PS^2}}}
{{{QS^2=9^2+12^2=81+144=225}}} and {{{QS=sqrt(225)=15}}} .
 
In quadrilateral {{{PQRS}}} ,
vertex {{{R}}} is adjacent to vertices {{{Q}}} and {{{S}}} .
Since there is a right angle at {{{R}}} ,
triangle {{{RQS}}} is a right triangle with hypotenuse {{{QS=15}}} ,
and legs {{{QR=10}}} and {{{PS}}} .
Using the Pytahagorean theorem, we can find leg {{{PS}}} :
{{{QS^2=QR^2+PS^2}}}
{{{225=10^2+PS^2}}}--->{{{225=100+PS^2}}}--->{{{225-100=PS^2}}} .
So, {{{PS^2=125}}}--->{{{PS=sqrt(125)}}}}--->{{{highlight(PS=5sqrt(5))}}} .
If you want an approximate measure, {{{highlight(PS=about11.18)}}} (rounded).
 
{{{drawing(300,300,-1,13,-1,13,
triangle(0,0,12,0,0,9),
triangle(9.805,10.96,12,0,0,9),
rectangle(0,0,1,1),
line(8.715,10.74,8.935,9.65),
line(10.025,9.87,8.935,9.65),
locate(-0.2,0,P),locate(11.8,0,S),
locate(-0.2,9.7,Q),locate(9.6,11.7,R)
)}}}