What is the distance between point E and square ABCD?
ABCD is a square with an area of 1440 cm^2. It and point E forms a pyramid. Triangle ABE, triangle BCE, triangle CDE and triangle ADE are triangles (not equilateral triangles) with areas of 840 cm^2. Line AE, line BE, line CE and line DE have the same length. What is the distance between point E and square ABCD in cm?
The answer is 40.
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1 solution
The side length of square ABCD is sqrt(1440) cm.
The heights of the triangles ABE
=840*2/[sqrt(1440)]
=1680/[sqrt(1440)]
The length of line AE
=sqrt { {[sqrt(1440)]/2}^2+{1680/[sqrt(1440)]}^2 }
=sqrt(360+1960)
=sqrt(2320)
The length of its diagonal is
sqrt{ [sqrt(1440)]^2+[sqrt(1440)]^2}
=sqrt(1440+1440)
=sqrt(2880).
Imagine the centre of the square is a point : Ｆ
Let x be the distance between ABCD and E=distance between E and F
Lines AF's length are half of the length of the diagonals of square ABCD
Line AE's length:
AF^2+x^2=AE^2
{[sqrt(2880)]/2}^2+x^2=[sqrt(2320)]^2
720+x^2=2320
x^2=2320-720
x^2=1600
x=sqrt(1600)
x=40
So, the distance between point E and square ABCD is 40cm.
Used knowledge:
Pythagorean theorm: a^2+b^2=c^2
c=sqrt(a^2+b^2)