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11 May 2021, 06:30
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The figure above was cut out of paper and folded to form a pyramid with a square base and four equilateral triangles. If the side of the pyramid's base is 1, what is the height of the pyramid?
A. 1/2
B. √2/2
C. (√3)/2
D. (√5)/2
E. √3
Attachment:
1.png [ 2.2 KiB | Viewed 2526 times ]
21 May 2021, 08:26
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median of the triangle will act as the hypotenuse of the pyramid
median of the triangle=\sqrt{1^2-(1/2)^2} = \sqrt{3}/2
height of the pyramid= \sqrt{([square_root]3}/2)^2-(1/2)^2[/square_root] = 1/\sqrt{2} = \sqrt{2}/2
median of the triangle=\sqrt{1^2-(1/2)^2} = \sqrt{3}/2
height of the pyramid= \sqrt{([square_root]3}/2)^2-(1/2)^2[/square_root] = 1/\sqrt{2} = \sqrt{2}/2
25 May 2021, 00:24
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The pyramid will be formed when the 4 equilateral triangles are folded over and converge at the top vertex/apex of the pyramid
Because we have a square base, each side of the equilateral triangle will be 1.
Then, from the top apex of the pyramid we can draw a perpendicular height down to the center of the square. Call this Height of the Pyramid = H
This will fall exactly in the geometric center of the figure ——-> which is (1/2) from the side of the square base.
The lateral “slant height” of the pyramid will form a right triangle with the (1/2) side and the perpendicular height of the pyramid. The right triangle will be the following:
——-The lateral “slant height” will be the hypotenuse of this right triangle.
——The legs of the right triangle will be:
-the (1/2) distance from the side of the square base and
-the Perpendicular Height of the pyramid we called (H)
Because these are equilateral triangles and we know the side lengths, the “slant height” on the outside of the pyramid will be the same thing as the Height of the equilateral triangle, were it laying flat as a 2-D object.
Height of an Equilateral Triangle = (side) * sqrt(3) * (1/2)
Thus: lateral “slant height” of pyramid = (1) * sqrt(3) * (1/2) = (sqrt(3)) / (2)
We can then use the Pythagorean Theorem to solve for the height of the pyramid:
(H)^2 + (1/2)^2 = [ sqrt(3) / 2 ]^2
(H)^2 + (1/4) = (3/4)
(H)^2 = 2/4
taking square root of each side
H = sqrt(2) / 2
B
Posted from my mobile device
Because we have a square base, each side of the equilateral triangle will be 1.
Then, from the top apex of the pyramid we can draw a perpendicular height down to the center of the square. Call this Height of the Pyramid = H
This will fall exactly in the geometric center of the figure ——-> which is (1/2) from the side of the square base.
The lateral “slant height” of the pyramid will form a right triangle with the (1/2) side and the perpendicular height of the pyramid. The right triangle will be the following:
——-The lateral “slant height” will be the hypotenuse of this right triangle.
——The legs of the right triangle will be:
-the (1/2) distance from the side of the square base and
-the Perpendicular Height of the pyramid we called (H)
Because these are equilateral triangles and we know the side lengths, the “slant height” on the outside of the pyramid will be the same thing as the Height of the equilateral triangle, were it laying flat as a 2-D object.
Height of an Equilateral Triangle = (side) * sqrt(3) * (1/2)
Thus: lateral “slant height” of pyramid = (1) * sqrt(3) * (1/2) = (sqrt(3)) / (2)
We can then use the Pythagorean Theorem to solve for the height of the pyramid:
(H)^2 + (1/2)^2 = [ sqrt(3) / 2 ]^2
(H)^2 + (1/4) = (3/4)
(H)^2 = 2/4
taking square root of each side
H = sqrt(2) / 2
B
Posted from my mobile device
