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Chapter 2: Problem 54

Compute the volume of a parallelepiped all of whose faces are congruent rhombi with the side \(a\) and an angle \(60^{\circ}\).

Short Answer

Expert verified
The volume of the parallelepiped is \( \frac{a^3 \sqrt{3}}{2} \).

Step by step solution

01

Understand the Shape

A parallelepiped is a six-faced figure (a polyhedron) where each face is a parallelogram. If all faces are congruent rhombi with side length \( a \) and an angle of \( 60^{\circ} \), it means all faces are identical rhombi.
02

Diagonal Properties of a Rhombus

A rhombus with one angle \( 60^{\circ} \) will have diagonals that bisect each other at right angles. We can use the cosine rule: if all sides \( a \) form a \( 60^{\circ} \) angle, one diagonal is \( a \) and the adjacent angle is \( 120^{\circ} \). The longer diagonal can be computed using \( a^2 = d_1^2 + d_2^2 - 2d_1d_2\cos(60^\circ) \) with \( d_1 = a \).
03

Calculate the Height of the Parallelepiped

The height of the parallelepiped is given by the perpendicular distance between the parallel rhombus faces. For rhombi, the perpendicular height (h) from a side to the opposite parallel side can be computed using the area of the rhombus formula: \( \text{area} = a^2 \sin(60^{\circ}) = \frac{a^2 \sqrt{3}}{2} \). The height is a part of the volume calculation, yet it's included here for computing parameters.
04

Determine the Volume Formula

Since the base is a rhombus with area \( \frac{a^2 \sqrt{3}}{2} \) and height is the perpendicular distance derived from that area, the volume of the parallelepiped is given by the product of the base area and the height along one grid stack: \( V = \text{Area of rhombus} \times h = a^3 \sqrt{3} \times 1/2 \).
05

Compute the Volume

Substitute the values into the volume formula. The volume \( V \ \) of the parallelepiped is calculated as: \[V = \frac{a^3 \sqrt{3}}{2}.\] This is derived under the assumption of one particular alignment allowing for interior grid staking of rhombic faces as 'dual axis' opposite faces would evaluate this structure along axes defined by different parallelogram interfaces.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Congruent Rhombi
Congruent rhombi are geometric figures that have identical size and shape. A rhombus is a type of parallelogram where all sides are equal in length and opposite angles are equal. When discussing congruent rhombi, we mean that every face of the geometry in question is the same rhombus. In our exercise, all faces of the parallelepiped are congruent rhombi. This means each face has side length \( a \) and angles measuring \( 60^{\circ} \) and \( 120^{\circ} \).
Understanding congruent rhombi helps in knowing that all dimensions of these faces are the same. This uniformity simplifies calculations, such as finding the volume of a more complex shape like a parallelepiped.
Rhombi's special properties such as equal sides and bisecting diagonals are key in geometry problem solving and derive from the basic features of parallelograms. Let's see how these properties assist in geometry problem solving.
Geometry Problem Solving
Geometry problem solving often involves understanding the properties of the shapes involved. With our exercise, a problem arises: finding the volume of a parallelepiped. Initially, it might seem daunting due to its three-dimensionality and angled faces.
The solution begins with grasping the parallels between the identical rhombi that form all faces of this shape. Each rhombus face shares certain properties, allowing us to use formulas specific to these properties to solve the problem.
  • Each rhombus is defined by its sides and angles, making it easier to compute the area, which we use as the base of the parallelepiped.
  • The perpendicular height across parallel rhombus faces is determined using trigonometric functions, which reveal how distance varies relative to the orientation of the shape.
As we solve the problem, we follow the logical steps of breaking down the structure into understandable formulas鈥攖he base area and height. By mastering these steps, problem solving in geometry becomes systematic, aligning problems to known solutions or formulas.
Trigonometric Functions in Geometry
Trigonometric functions are indispensable tools in geometry, especially in solving problems involving angles and distances. In the exercise, trigonometry helps calculate the area of the rhombus faces of the parallelepiped from its angle of \( 60^{\circ} \).
Utilizing the sine function, we find the area of every rhombus: \( \text{area} = a^2 \sin(60^{\circ}) = \frac{a^2 \sqrt{3}}{2} \). This knowledge is crucial for determining the height and ultimately the volume of the parallelepiped.
The interdependence between the dimensions of the rhombi and their orientation via trigonometry unveils complex spatial relationships in a seemingly simple way. Recognizing that a \( 60^{\circ} \) angle can help calculate perpendicular dimensions opens avenues to various geometric calculations. Trigonometric functions bridge the gap between abstract geometric angles and measurable distances, thereby simplifying complex shapes.

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Most popular questions from this chapter

Classification of regular polyhedra. Let us take into account that a convex polyhedral angle has at least three plane angles, and that their sum has to be smaller than \(4 d(\S 48)\). Since in a regular triangle, every angle is \(\frac{2}{3} d\), repeating it 3,4 , or 5 times, we obtain the angle sum smaller than \(4 d\), but repeating it 6 or more times, we get the angle sum equal to or greater than \(4 d\). Therefore convex polyhedral angles whose faces are angles of regular triangles can be of only three types: trihedral, tetrahedral, or pentahedral. Angles of squares and regular pentagons are respectively \(d\) and \(\frac{6}{5} d\). Repeating these angles three times, we get the sums smaller than \(4 d\), but repeating them four or more times, we get the sums equal to or greater than \(4 d\). Therefore from angles of squares or regular pentagons, only trihedral convex angles can be formed. The angles of regular hexagons are \(\frac{4}{3} d\), and of regular polygons with more than 6 sides even greater. The sum of three or more of such angles will be equal to or greater than \(4 d\). Therefore no convex polyhedral angles can be formed from such angles. It follows that only the following five types of regular polyhedra can occur: those whose faces are regular triangles, meeting by three, four or five triangles at each vertex, or those whose faces are either squares, or regular pentagons, meeting by three faces at each vertex.

Compute the angle between diagonals of two adjacent faces of a cube. (Consider first the diagonals that meet, then skew ones.)

Let three figures \(\Phi, \Phi^{\prime}\), and \(\Phi^{\prime \prime}\) be symmetric: \(\Phi\) and \(\Phi^{\prime}\) about a plane \(P\), and \(\Phi^{\prime}\) and \(\Phi^{\prime \prime}\) about a plane \(Q\) perpendicular to \(P\). Prove that \(\Phi\) and \(\Phi^{\prime \prime}\) are symmetric about the intersection line of \(P\) and \(Q\).

Determine the number of planes of symmetry of a regular pyramid with \(n\) lateral faces. 131\. Let three figures \(\Phi, \Phi^{\prime}\), and \(\Phi^{\prime \prime}\) be symmetric: \(\Phi\) and \(\Phi^{\prime}\) about a plane \(P\), and \(\Phi^{\prime}\) and \(\Phi^{\prime \prime}\) about a plane \(Q\) perpendicular to \(P\). Prove that \(\Phi\) and \(\Phi^{\prime \prime}\) are symmetric about the intersection line of \(P\) and \(Q\). 132\. What can be said about the figures \(\Phi\) and \(\Phi^{\prime \prime}\) of the previous problem if the planes \(P\) and \(Q\) make the angle: (a) \(60^{\circ} ?\) (b) \(45^{\circ} ?\)

Compute the total surface area of the tetrahedron all of whose edges have the same length \(a\).
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