Answer:
The distance is: [tex]\sqrt3\ cm\approx1,73\,cm[/tex]
Step-by-step explanation:
The distance of point E to the PQR plane it is the hight (vertical) of piramid PRQE
If point P is located in the middle of line EH, point Q is in the middle of line EF, and point R is in the middle of line AE than:
EP = EQ = ER = 0.5EF = 3 cm and m∠REQ = m∠QEP = m∠REP = 90° so triangles RQE, QPE and PRE are congruent.
RQ = QP = PR so triangle PQR is equilateral and from Pythagorean theorem (for ΔRQE):
[tex]RQ^2=ER^2+EQ^2=3^2+3^2=2\cdot3^2\ \ \implies\ \ RQ=3\sqrt2[/tex]
Then: [tex]RN=\dfrac{RQ\,\sqrt3}2[/tex]
and: [tex]RK=\dfrac23RN=\dfrac{RQ\,\sqrt3}3=\dfrac{3\sqrt2\cdot\,\sqrt3}3=\sqrt6[/tex]
Therefore from Pythagorean theorem (for ΔERK):
[tex]EK^2+RK^2=ER^2\\\\EK^2=ER^2-RK^2\\\\EK^2=3^2-(\sqrt6)^2\\\\EK^2=9-6=3\\\\EK=\sqrt3\ cm\approx1,73\,cm[/tex]
