 5.1: How do you find the measure of one exterior angle of a regular poly...
 5.2: How can you find the number of sides of an equiangular polygon by m...
 5.3: How do you construct a rhombus by using only a ruler or doubleedge...
 5.4: How do you bisect an angle by using only a ruler or doubleedged st...
 5.5: How can you use the converse of the Rectangle Diagonals Conjecture ...
 5.6: How can you use the Triangle Midsegment Conjecture to find a distan...
 5.7: Find x and y.
 5.8: The perimeter is 266 cm. Find x.
 5.9: Find a and c.
 5.10: MS is a midsegment. Find the perimeter of MOIS.
 5.11: Find x.
 5.12: Find y and z.
 5.13: Copy and complete the table below by placing a yes (to mean always)...
 5.14: Application A 2inchwide frame is to be built around the regular d...
 5.15: Developing Proof Find the measure of each lettered angle. Explain h...
 5.16: Archaeologist Ertha Diggs has uncovered one stone that appears to b...
 5.17: Kite ABCD has vertices A(3, 2), B(2, 2), C(3, 1), and D(0, 2). Find...
 5.18: When you swing left to right on a swing, the seat stays parallel to...
 5.19: Construction The tiling of congruent pentagons shown below is creat...
 5.20: Construction An airplane is heading north at 900 km/h. However, a 5...
 5.21: Construction In Exercises 2124, use the given segments and angles t...
 5.22: Construction In Exercises 2124, use the given segments and angles t...
 5.23: Construction In Exercises 2124, use the given segments and angles t...
 5.24: Construction In Exercises 2124, use the given segments and angles t...
 5.25: Three regular polygons meet at point B. Only four sides of the thir...
 5.26: Find x
 5.27: Developing Proof Prove the conjecture: The diagonals of a rhombus b...
 5.28: Developing Proof Use a labeled diagram to prove the Parallelogram O...
Solutions for Chapter 5: Discovering and Proving Polygon Properties
Full solutions for Discovering Geometry: An Investigative Approach  4th Edition
ISBN: 9781559538824
Solutions for Chapter 5: Discovering and Proving Polygon Properties
Get Full SolutionsChapter 5: Discovering and Proving Polygon Properties includes 28 full stepbystep solutions. This expansive textbook survival guide covers the following chapters and their solutions. Discovering Geometry: An Investigative Approach was written by and is associated to the ISBN: 9781559538824. This textbook survival guide was created for the textbook: Discovering Geometry: An Investigative Approach, edition: 4. Since 28 problems in chapter 5: Discovering and Proving Polygon Properties have been answered, more than 22174 students have viewed full stepbystep solutions from this chapter.

Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x  x) (x  x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

Free variable Xi.
Column i has no pivot in elimination. We can give the n  r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Inverse matrix AI.
Square matrix with AI A = I and AAl = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B1 AI and (AI)T. Cofactor formula (Al)ij = Cji! detA.

Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

Nilpotent matrix N.
Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b  Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) 1 AT.

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.

Toeplitz matrix.
Constant down each diagonal = timeinvariant (shiftinvariant) filter.

Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and AI are BT AT and (AT)I.

Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II·

Unitary matrix UH = U T = UI.
Orthonormal columns (complex analog of Q).

Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn 1 with P(Xi) = bi. Vij = (Xi)jI and det V = product of (Xk  Xi) for k > i.