Answer:
HI = [tex]4\sqrt{41}[/tex] , IJ = [tex]4\sqrt{41}[/tex]
Step-by-step explanation:
The number of AB spaces is 15, the BC is 20, and the CD is 15. Looking at the given length, you can see that it is twice the number of cells.
If you draw a line perpendicular to the line HG in I, it becomes a right triangle. The lengths of HI and IJ can be calculated using Pythagoras on a right triangle.
Suppose that the vertical line is L.
HL = 10
IL = 8
(HI)² = (HL)²+ (IL)² = 10²+8²=164
HI = [tex]2\sqrt{41\\[/tex] -> (It's twice the number of spaces.) -> HI = [tex]4\sqrt{41}[/tex]
IJ can be obtained in the same way.
IJ = [tex]2\sqrt{41\\[/tex] -> (It's twice the number of spaces.) -> IJ = [tex]4\sqrt{41}[/tex]
