Answer:
[tex]SU=3.5\ units[/tex]
Step-by-step explanation:
step 1
In the isosceles right triangle PQT
PQ=PT ----> because is an isosceles triangle
Applying the Pythagoras Theorem
we have
[tex]PQ=PT =\sqrt{2}\ units[/tex]
[tex]QT^{2}=PQ^{2} +PT^{2}[/tex]
substitute the values
[tex]QT^{2}=\sqrt{2}^{2} +\sqrt{2}^{2}[/tex]
[tex]QT^{2}=4[/tex]
[tex]QT=2\ units[/tex]
step 2
In the square QRST
[tex]RS=QT=2\ units[/tex]
step 3
In the right triangle RSU
Applying the Pythagoras Theorem
[tex]RU^{2}=RS^{2} +SU^{2}[/tex]
we have
[tex]RU=4\ units[/tex]
[tex]RS=2\ units[/tex]
substitute the values and solve for SU
[tex]4^{2}=2^{2} +SU^{2}[/tex]
[tex]SU^{2}=4^{2}-2^{2}[/tex]
[tex]SU^{2}=12[/tex]
[tex]SU=2\sqrt{3}\ units[/tex]
[tex]SU=3.5\ units[/tex]
