   Chapter 9.2, Problem 27E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# For a regular square pyramid, suppose that the altitude has a measure equal to that of each edge of the base. If the volume of the pyramid is 72 i n 3 , find the total area of the pyramid. To determine

To find:

The total area of the pyramid.

Explanation

Given:

A regular square pyramid as below,

and the altitude has a measure equal to that of each edge of the base i.e. h=x. The volume of the pyramid is 72in3.

Properties used:

A pyramid is made by connecting a base to an apex. The base is flat with straight edges, no curves, hence, a polygon and all other faces are triangles.

A regular pyramid is a pyramid whose base is a regular polygon and whose lateral edges are all congruent.

The lateral area L of a regular pyramid with slant height of length l and perimeter P of the base is given by

L=12lP.

The total surface area T of a pyramid with lateral area L and base area B is given by

T=L+B.

According to the Pythagorean theorem, in a right-angled triangle

hypotenuse2=base2+perpendicular2.

In a regular pyramid, the lengths of the apothem a of the base, the altitude h, and the slant height l satisfy the Pythagorean Theorem; that is, l2=a2+h2.

The perimeter of the square is four times the side.

The volume V of a pyramid having a base area B and an altitude of length h is given by V=13Bh.

Calculation:

From the given figure

The sides of the square base measure x each and the apothem is half the base length. Hence, a=x2. The volume is 72in3.

The area of the square base B=x2.

Since, the volume V of a pyramid having a base area B and an altitude of length h is given by V=13Bh. Substituting values, we get

72=13x2x72=x33x3=72×3x=6in

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