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The Great Pyramid of Cheops in Egypt has a square base with side length of 800 ft. It's triangular faces are equilateral.


What I Can Do The Great Pyramid Of Cheops In Egypt Has A Square Base With Side Length Of 800 Ft Its Triangular Faces Are Equilateral class=

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Answer:

the height of each triangle is 400[tex]\sqrt{3}[/tex]ft or 692.82 ft

the height of the pyramid is 400[tex]\sqrt{2}[/tex]ft or 565.6854 ft

Step-by-step explanation:

the height of each triangle can be determined using the Phytagorean Theorem or the 30-60-90 Special Right Triangle relation,

since the square base has a side measuring 800 ft, and observe that the base of the triangle is a side of the square, so obviously the base of the triangle measures 800ft. Since the triangle is equilateral as stated, then all the sides of the triangle measures 800 ft. If you are going to split the triangle in half, you'll bisect the 60° in the vertex and the base, so you'll have a 30-60-90 triangle with base 400 ft., the side opposite the 60° angle is √3 times the measure of the side opposite the 30° angle, the side opposite the 30° is 400ft, so basically, the height or the side opposite the 60° is 400√3.

The height of the pyramid can be solved by:

if you are going to put the figure together to form the pyramid, observe that the height of the triangles is the slant height of the pyramid(not the height), just look the pyramid in side view and you'll observe that you can form a right triangle with the base (b) half the square base ,

hypotenuse as the slant height (l) and the unknown height (h)[tex]l^{2} = h^{2} + b^{2} \\-h^{2} =b^{2}-l^{2} \\h^{2} =l^{2} -b^{2} \\h^{2} =(400\sqrt{3} )^{2} - (400)^{2} \\h^{2} =(400)^{2} (3)-(400)^{2} \\h^{2} =(400)^{2} (3-1)\\h^{2} =(400)^{2}(2)\\ h^{} =\sqrt{(400)^{2}(2) } \\h= 400\sqrt{2}[/tex], just use the phytagorean theorem,