Answer:
[tex]\text{8.93m}[/tex]
Step-by-step explanation:
[tex]\text{Please see the labeled diagram I posted.}[/tex]
[tex]\text{Here, we have to find the height }h\text{, which is the shortest distance between}\\\text{the parallel sides.}\\\text{Given, AC = BD}\\\therefore\ \text{AC = }2y-6\\\text{And,}\\\dfrac{x}{2}+7=2y-6\\\\ \text{or, }x=4y-26......(1)[/tex]
[tex]\text{Also,}\\\text{Perimeter of trapezium = 47m}\\\text{or, AB + BD + CD + AC = 47}\\\text{or, 9 + }2y-6+3x+2y-6=47\\\text{or, }3x+4y-3=47\\\text{or, }3x+4y=50\\\text{Using equation(1),}\\3(4y-26)+4y=50\\\text{or, }12y-78+4y=50\\\text{or, }16y=128\\\text{or, }y=8[/tex]
[tex]\text{Here, triangles ACD and BFD are congruent triangles by S.S.S. axiom. }\\\text{Hence, CE and FD are equal since they are corresponding sides of congruent}\\\text{triangles.}\\\therefore\ \text{CD = CE + EF + FD}\\\Rightarrow\ 3x\text{ = 2FD + 9}[/tex]
[tex]\text{From equation(1),}\\3(4y-26)=2\text{FD}+9\\\text{or, }12y-78-9=2\text{FD}\\\text{or, }2\text{FD}=12(8)-78-9=9\\\text{or, FD}=9/2\text{m}[/tex]
[tex]\therefore\ \text{BD = }2y-6=2(8)-6=10\text{m}\\\text{Using pythagoras theorem in triangle BFD,}\\\text{BF}=\sqrt{\text{BD}^2-\text{FD}^2}=\sqrt{10^2-(9/2)^2}\approx8.93\text{m}[/tex]
[tex]\text{So the distance between the two parallel sides is approximately 8.93m.}[/tex]
