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 apex, or top of the pyramid, to each vertex of the base?.

A three-dimensional figure is a pyramid. Its base is a flat polygon. The remaining faces are all triangles and are referred to as lateral faces. The number of sides on its base is equal to the number of lateral faces. The line segments that two faces intersect to form its edges. The intersection of three or more edges forms a vertex. All the faces, with the exception of the base, join at the apex, a vertex at the top. The base's shape is given by the apex, which is located in opposition to it. In the right pyramid, the apex is precisely over the center of the base. The center of the base will be where a perpendicular line from the apex to the base intersects.

Calculation

So in the question they mentioned Altitude=10 and Base =6.

So using pythagorean theorem,

x2=y2+(10)2-eq1

with y half of the length of the diagonal of the base.

2y=length of the diagnol

and (2y)2 = 72

⇒4y2=72

∴y = 3√2.

So substitute y in eq1 and get x.

⇒ ( 3√2)2+(10)2=(x)2

⇒(x)2=118

∴x=10.86.

x≈10.9

We need to use the properties of Pyramid and Right Angled Triangle in this question. The distance from apex to each vertex is 10.9.

To refer more about Pyramid and Right Angled Triangle , visit:

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