A right square pyramid has an altitude of 10 and each side of the base is 6. To the nearest tenth of a centimeter, what is the distance from the top of the pyramid, to each vertex of the base?
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Answer:
10.9
Step-by-step explanation:
I just used a pyramid calculator online and inputed the base and height.
The distance from the top of the pyramid, to each vertex of the base would be 10.5 unit.
If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:
[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]
where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).
A right square pyramid has an altitude of 10 and each side of the base is 6. we need to find the slant height of the pyramid.
From the Pythagoras' theorem
[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]
[tex]s^2 = h^2 + (b/2)^2\\\\s^2 = 10^2 + (6/2)^2\\\\s^2 = 100 + 9\\\\s^2 = 109\\\\s = \sqrt109\\\\s = 10.5[/tex]
Thus, the distance from the top of the pyramid, to each vertex of the base would be 10.5 unit.
Learn more about Pythagoras' theorem here:
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