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Chapter 4: Problem 35

Show that if the diagonals of a parallelogram are perpendicular, it is necessarily a rhombus.

Short Answer

Expert verified
If the diagonals of a parallelogram are perpendicular, it has equal sides; hence, it's a rhombus.

Step by step solution

01

Understand the properties of a parallelogram

A parallelogram is a quadrilateral with opposite sides parallel and equal in length. It also has opposite angles equal and the diagonals bisect each other.
02

Describe the diagonals in a rhombus

In a rhombus, which is a special type of parallelogram, all sides are equal in length. The diagonals of a rhombus are perpendicular and they bisect each other at right angles.
03

Set up the condition for perpendicular diagonals

Suppose diognals of a parallelogram are \( AC \) and \( BD \) in which \( AC \perp BD \), indicating they intersect at a right angle. Let point \( O \) be the intersection.
04

Use vector analysis to show all sides are equal

If we consider position vectors \( \vec{a} \) and \( \vec{b} \) for any two adjacent sides of a parallelogram, then diagonals can be represented in vectors as \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \). If \( \vec{a} + \vec{b} \) is perpendicular to \( \vec{a} - \vec{b} \), their dot product must be zero: \ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 0 \ This implies that \( \| \vec{a} \|^2 = \| \vec{b} \|^2 \), meaning \( \| \vec{a} \| = \| \vec{b} \| \) showing all sides of the parallelogram are equal.
05

Conclude that the parallelogram is a rhombus

Since we've shown that the condition of perpendicular diagonals leads to all sides being equal in length, the initial parallelogram must be a rhombus, given that a rhombus is a parallelogram with all sides equal.

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Key Concepts

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

Parallelogram
A parallelogram is a popular shape in geometry. It has some unique properties that make it very special. A quadrilateral becomes a parallelogram when both pairs of opposite sides are parallel. Here are some important facts about parallelograms:
  • Opposite sides are not only parallel but also equal in length.
  • It has another special property where the opposite angles are equal.
  • The diagonals in a parallelogram bisect each other, meaning each diagonal cuts the other into two equal parts.
The properties of parallelograms are foundational for understanding more specific types of quadrilaterals, like rectangles, rhombi, and squares. Understanding these basics paves the way for learning more complex shapes.
Rhombus
A rhombus is a fascinating shape in geometry. While it shares many properties with a parallelogram, it also has its unique characteristics:
  • All four sides of a rhombus are equal in length, which is the most defining feature.
  • Like all parallelograms, it has opposite sides that are both equal and parallel.
  • Additionally, opposite angles in a rhombus are equal.
One of the key features of a rhombus is its diagonals. They are special compared to those of a standard parallelogram:
  • The diagonals in a rhombus are perpendicular, meaning they intersect at a 90-degree angle.
  • Each diagonal bisects the other, dividing the rhombus into four right-angle triangles.
These properties are what make a rhombus not just any parallelogram but a special one! When a parallelogram's diagonals are perpendicular, it assures that all sides are equal, making it a rhombus.
Diagonals
In geometry, diagonals are the lines connecting opposite corners of a shape. They play a crucial role in understanding the properties of polygons, especially quadrilaterals. Let's focus on diagonals within parallelograms and rhombi:
  • In a parallelogram, the diagonals bisect each other, meaning they split each other into two equal segments.
  • This bisecting property holds true for all parallelograms, but what about when the diagonals intersect perpendicularly?
When diagonals in a parallelogram are perpendicular, they signal something special: the parallelogram is actually a rhombus. Here’s why:
  • The perpendicular nature of the diagonals indicates all sides are equal. This is due to the fact that the diagonals create right triangles, where the hypotenuse (the sides of the rhombus) are all equal.
  • Mathematically, this is derived using vector properties where if diagonal vectors are perpendicular, their dot product equals zero, proving all sides are equal.
Understanding diagonals isn't just about connecting corners; it's about revealing hidden properties of the shapes they belong to.

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Most popular questions from this chapter

Find the point of intersection (if any) of the following pairs of lines. $$ \text { a. } \begin{array}{ll} x=3+t & x=4+2 s \\ & y=1-2 t & y=6+3 s \\ & z=3+3 t & z=1+s \\ & x=1-t & x=2 s \end{array} $$ b. \(\quad y=2+2 t \quad y=1+s\) $$ \begin{array}{c} z=-1+3 t \quad z=3 \\ \text { c. }\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 3 \\ -1 \\ 2 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right]+s\left[\begin{array}{l} 2 \\ 0 \\ 3 \end{array}\right]} \end{array} $$ d. \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{r}4 \\ -1 \\\ 5\end{array}\right]+t\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]\) \(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}2 \\\ -7 \\ 12\end{array}\right]+s\left[\begin{array}{r}0 \\ -2 \\\ 3\end{array}\right]\)

Let \(\mathbf{v} \neq \mathbf{0}\) be a nonzero vector and let \(a \neq 0\) be a scalar. If \(\mathbf{u}\) is any vector, show that the projection of \(\mathbf{u}\) on \(\mathbf{v}\) equals the projection of \(\mathbf{u}\) on \(a \mathbf{v}\).

Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be pairwise orthogonal vectors. a. Show that \(\|\mathbf{u}+\mathbf{v}+\mathbf{w}\|^{2}=\|\mathbf{u}\|^{2}+\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}\). b. If \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are all the same length, show that they all make the same angle with \(\mathbf{u}+\mathbf{v}+\mathbf{w}\).

Show that the triangle with vertices \(A(4,-7,9), B(6,4,4),\) and \(C(7,10,-6)\) is not a rightangled triangle.

a. Show that \(\mathbf{w} \cdot(\mathbf{u} \times \mathbf{v})=\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=\mathbf{v} \times(\mathbf{w} \times \mathbf{u})\) holds for all vectors \(\mathbf{w}, \mathbf{u},\) and \(\mathbf{v}\) b. Show that \(\mathbf{v}-\mathbf{w}\) and \((\mathbf{u} \times \mathbf{v})+(\mathbf{v} \times \mathbf{w})+(\mathbf{w} \times \mathbf{u})\) are orthogonal.
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