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2 years ago

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The Palace of Peace and Accord in Astana, Kazakhstan, was built in 2006. The building has a square right pyramid structure. The

length of a side of the square base is 62 m, and the vertical height of the pyramid is also 62 m. What is the height, h, of one of the triangular faces? Round your answer to the nearest tenth.

2 answers:

Molodets [167]2 years ago

7 0

Answer: 69.32 m

Step-by-step explanation:

Since, slant height of the square right pyramid, $s^2=\sqrt{h^2+(1/4) a^2}$

where,

h = height of the pyramid

a= side of the square

And, here h= 62 m and a= 62 m

Therefore, $s^2=\sqrt{62^2+(1/4) 62^2}$

⇒ $s^2=\sqrt{3844+961}$

⇒ $s^2=\sqrt{4805}$

⇒s=69.3181073025≈ 69.32 m

Thus, the height of triangular face = 69.32 ( approx)

dem82 [27]2 years ago

3 0

The height you are looking for is called the "slant height".
slant height^2 = height^2 + (half the base length)^2
slant height^2 = 62^2 + 31^2
slant height^2 = 3,844 + 961
slant height^2 = 4,805
slant height = <span> <span> <span> 69.3181073025 </span> </span> </span>
slant height = <span> <span> <span> 69.3 (rounded)</span></span></span>

Source:
http://www.1728.org/volpyrmd.htm

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