Maria is setting up a geocaching course. She begins at point \(A\), then walks south \(150\) m to point \(B\), then walks east \(140\) m to point \(C\), and then walks north \(150\) m to point \(D\).
She hides a final cache at point \(E\), which is located \(50\) m east of path \(AB\) and \(30\) m north of path \(BC\), as shown in the diagram.
Determine the distance from \(A\) to \(E\), the distance from \(D\) to \(E\), and the total distance around the five-sided route \(ABCDEA\).
Note: You may find the following useful:
The Pythagorean Theorem states, "In a right-angled triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides."
In the right-angled triangle shown, \(c\) is the hypotenuse, \(a\) and \(b\) are the lengths of the other two sides, and \(c^2=a^2+b^2\).
Let \(F\) be the point on \(AB\) with \(FE
=50\) m. Let \(H\) be the point
on \(BC\) with \(EH=30\) m.
Extend \(EF\) to \(G\) on \(CD\).
Since \(ABCD\) is a rectangle and \(FG\) is perpendicular to \(AB\), then \(FG\) is perpendicular to \(CD\) and \(FGCB\) is a rectangle. Therefore, \(FB= EH=GC=30\) m.
Also, \(EG = BC - FE = 140 - 50 = 90\)
m, and \(DG = AF = AB - FB = 150 - 30 =
120\) m.
Since \(\triangle AFE\) and \(\triangle DGE\) are right-angled triangles, we can use the Pythagorean Theorem to determine the lengths of \(AE\) and \(DE\).
In \(\triangle AFE\), \(AE^2=AF^2+FE^2
=120^2+50^2
=14\,400+2500
=16\,900\).
Thus, \(AE=130\), since \(AE > 0\).
In \(\triangle DGE\), \(DE^2=DG^2+EG^2
=120^2+90^2
=14\,400+8100
=22\,500\).
Thus, \(DE=150\), since \(DE > 0\).
Therefore, \(AE=130\) m and \(DE=150\) m.
Also, the perimeter of pentagon \(ABCDE\) is equal to \(AB + BC + CD + DE + AE = 150 + 140 + 150 + 150 +
130= 720\) m.