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11. Solving Problems in the Coordinate Plane
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United States of America
Grades 9-12

11.03 Classifying polygons in the coordinate plane

Lesson
Practice

Introduction

In previous lessons, we have explored the properties of various triangles and quadrilaterals. In this lesson, we will learn how to use the slopes of the sides of a polygon to classify them precisely based on the properties of their sides and angles in the coordinate plane.

Classifying polygons in the coordinate plane

Polygons can be classified based on their side lengths and angle measures.

For triangles, we have the following classifications:

Equilateral triangle

A triangle with three sides of equal length and three 60\degree interior angles

A triangle with all 3 sides marked congruent and all 3 angles marked congruent and labeled 60 degrees
Isosceles triangles

A triangle containing at least two sides of equal length and two equal interior angle measures

A triangle with two sides marked congruent and the angles between the congruent sides and the third side are marked congruent
Scalene triangle

A triangle containing three unequal side lengths and three unequal angle measures

A triangle where one angle has a single angle marking, one has a double angle marking and one has a triple angle marking. One side has a single marking, one hase a double marking and one hase a triple marking
Right triangle

A triangle containing an interior right angle

Two right triangles: one having two congruent sides, and the other having no congruent sides.

Note that it is possible for a triangle to be a scalene right triangle or an isosceles right triangle.

Triangles that are not right triangles can be further classified as acute triangles (if all angle measures are smaller than 90 \degree) or obtuse triangles (if one angle measure is larger than 90 \degree).

For quadrilaterals, we have the following classifications:

Square

A quadrilateral with four right angles and four equal-length sides

A quadrilateral with 4 congruent sides, and 4 right angles.
Rectangle

A quadrilateral containing four right angles

A quadrilateral with 2 pairs of congruent sides, and 4 right angles.
Rhombus

A quadrilateral containing four congruent sides.

A quadrilateral with 4 congruent sides.
Parallelogram

A quadrilateral containing two pairs of parallel sides

A quadrilateral with 2 pairs of parallel sides.

We want to classify polygons as precisely as possible:

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For example, the triangle is a right-triangle and isosceles, so we can say it is a right-isosceles triangle to be as precise as possible.

The quadrilateral could be called a rectangle or a rhombus, but the most precise classification is a square.

If we know the coordinates of the vertices of a polygon, we can find the most precise classification using the lengths and slopes of line segments joining pairs of vertices.

  1. Find the slope of each side to determine if any sides are parallel or perpendicular.
  2. Find the length of each side to determine if any sides are congruent.

Examples

Example 1

Consider the given quadrilateral.

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a

Determine the slopes of \overline{AB}, \overline{BC}, \overline{CD}, and \overline{AD}.

Worked Solution
Create a strategy

The slopes of all the sides will help us to determine which sides are parallel and which are perpendicular.

Since the quadrilateral is displayed on the coordinate plane, we can read the rise and run off the coordinate plane instead of using the slope formula.

Apply the idea

Finding the slope of \overline{AB} :

\displaystyle \text{slope of }{\overline{AB}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle -\dfrac{3}{4}Substitute \text{rise}=-3 and \text{run}=4

Finding the slope of \overline{BC}:

\displaystyle \text{slope of }{\overline{BC}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle \dfrac{4}{3}Substitute \text{rise}=4 and \text{run}=3

Finding the slope of \overline{CD}:

\displaystyle \text{slope of }{\overline{CD}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle -\dfrac{3}{4}Substitute \text{rise}=-3 and \text{run}=4

Finding the slope of \overline{AD}:

\displaystyle \text{slope of }{\overline{AD}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle \dfrac{4}{3}Substitute \text{rise}=4 and \text{run}=3
Reflect and check

Notice that adjacent sides have opposite reciprocal slopes and that opposite sides have the same slopes.

b

Calculate the lengths of the sides.

Worked Solution
Create a strategy
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Since we have the diagram, we will use the Pythagorean theorem, but we could also substitute the coordinates of the vertices into the distance formula.

Apply the idea

Calculating AB:

\displaystyle AB^2\displaystyle =\displaystyle 4^2+3^2Substitute into Pythagorean theorem
\displaystyle AB^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle AB\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle AB\displaystyle =\displaystyle 5Evaluate the square root

Calculating BC:

\displaystyle BC^2\displaystyle =\displaystyle 3^2+4^2Substitute into Pythagorean theorem
\displaystyle BC^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle BC\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle BC\displaystyle =\displaystyle 5Evaluate the square root

Calculating AD:

\displaystyle AD^2\displaystyle =\displaystyle 3^2+4^2Substitute into Pythagorean theorem
\displaystyle AD^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle AD\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle AD\displaystyle =\displaystyle 5Evaluate the square root

Calculating DC:

\displaystyle DC^2\displaystyle =\displaystyle 4^2+3^2Substitute into Pythagorean theorem
\displaystyle DC^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle DC\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle DC\displaystyle =\displaystyle 5Evaluate the square root
Reflect and check

We only took the positive square root because we were looking for a length, which cannot be negative.

If you notice, the legs of the triangle with hypotenuse \overline{AB} are the same length as the legs of the triangle with hypotenuse \overline{DC}. To save some time and work, we could have used this to say AB=DC. We could have used similar reasoning to explain BC=AD.

c

Classify the quadrilateral as precisely as possible.

Worked Solution
Create a strategy

We will use the slopes and lengths found in parts (a) and (b) to determine if any sides are parallel, perpendicular, or congruent.

Apply the idea

From part (a), we can determine that all adjacent sides are perpendicular because their slopes are reciprocals with opposite signs. That is,

  • \overline{AB}\perp \overline{BC}
  • \overline{BC}\perp \overline{CD}
  • \overline{CD}\perp \overline{AD}
  • \overline{AD}\perp \overline{AB}

We can also determine that opposite sides are parallel because they have the same slope. That is,

  • \overline{AB}\parallel \overline{CD}
  • \overline{AD}\parallel \overline{BC}

From this, the quadrilateral is a rectangle or a square.

From part (b), we can determine that all 4 sides are congruent. That is,

  • \overline{AB}\cong \overline{BC} \cong \overline{CD} \cong \overline{AD}

From this, we can now classify the quadrilateral as a square.

Reflect and check

The shape is a quadrilateral, a parallelogram, a rectangle, and a rhombus, but the most precise classification is a square.

Example 2

Classify the triangle with vertices at A \left(-6,-2\right), B \left(-4,4\right), and C \left(14,-2\right) as precisely as possible.

Worked Solution
Create a strategy

Using technology or drawing a rough sketch can help us visualize the triangle, and that will help us come up with a strategy.

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From the sketch we can see that \overline{AB} and \overline{BC} appear to be perpendicular. It also appears that all the side lengths are different.

However, we must calculate the slopes and lengths of the sides to draw a correct conclusion.

Apply the idea

Find AB:

\displaystyle AB\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} Distance formula
\displaystyle =\displaystyle \sqrt{\left(-6-(-4)\right)^2+\left(-2-4\right)^2} Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(-2\right)^2+\left(-6\right)^2} Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{4+36} Evaluate the squares
\displaystyle =\displaystyle \sqrt{40} Evaluate the addition

Find BC:

\displaystyle BC\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} Distance formula
\displaystyle =\displaystyle \sqrt{\left(-4-14\right)^2+\left(4-\left(-2\right)\right)^2} Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(-18\right)^2+\left(6\right)^2} Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{324+36} Evaluate the squares
\displaystyle =\displaystyle \sqrt{360} Evaluate the addition

Find AC:

\displaystyle AC\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} Distance formula
\displaystyle =\displaystyle \sqrt{\left(-6-14\right)^2+\left(-2-\left(-2\right)\right)^2} Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(-20\right)^2+\left(0\right)^2} Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{400} Evaluate the squares
\displaystyle =\displaystyle 20 Evaluate the square root

None of the sides are congruent, so it is a scalene triangle.

Find the slope of \overline{AB}:

\displaystyle \text{slope of }{\overline{AB}}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle =\displaystyle \dfrac{4-\left(-2\right)}{-4-\left(-6\right)}Substitute the coordinates
\displaystyle =\displaystyle \dfrac{6}{2}Evaluate the subtraction
\displaystyle =\displaystyle 3Evaluate the division

Find the slope of \overline{BC}:

\displaystyle \text{slope of }{\overline{BC}}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle =\displaystyle \dfrac{-2-4}{14-\left(-4\right)}Substitute the coordinates
\displaystyle =\displaystyle \dfrac{-6}{18}Evaluate the subtraction
\displaystyle =\displaystyle -\dfrac{1}{3}Simplify the fraction

These two sides are perpendicular because their slopes are opposite reciprocals, so they form a right angle.

\triangle ABC is a right-scalene triangle.

Reflect and check

If \overline{AB} and \overline{BC} were not perpendicular, then we would also need to find the slope of \overline{AC} to check whether or not any two sides are perpendicular.

Example 3

Show that the quadrilateral with vertices at A\left(2,1\right),\,B\left(1,4\right),\,C\left(6,5\right),\, D\left(7,2\right) is not a rectangle.

Worked Solution
Create a strategy

By definition, a rectangle has 4 right angles, meaning the adjacent sides are all perpendicular to each other. If we find any two sides that are not perpendicular, then we have shown that the quadrilateral is not a rectangle.

Apply the idea

Slope of \overline{AB}:

\displaystyle m_{\overline{AB}}\displaystyle =\displaystyle \dfrac{4-1}{1-2}
\displaystyle =\displaystyle -3

Slope of \overline{BC}:

\displaystyle m_{\overline{BC}}\displaystyle =\displaystyle \dfrac{5-4}{6-1}
\displaystyle =\displaystyle \dfrac{1}{5}

These slopes are not opposite reciprocals, which means \overline{AB} is not perpendicular to \overline{BC}. This also means those sides do not form a 90\degree angle, so the quadrilateral is not a rectangle.

Reflect and check
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This sketch makes it easy to see that this figure does not have 4 right angles.

However, we could show that this figure has 2 sets of parallel sides, making it a parallelogram.

Example 4

Let ABCD be a rhombus with vertices at A\left(-2,-1\right),\, B\left(x,y\right),\, C\left(-2,-7\right) and D\left(-6,-4\right).

a

Find the coordinates of vertex B.

Worked Solution
Create a strategy

The sides of a rhombus are all the same length, so we need to find the length of one side first. We can do this by counting the rise (length of the vertical distance) and the run (length of the horizontal distance) from point C to point D. The rise and run from point C to B will have to be the same.

Apply the idea
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The horizontal distance from C to D is 4. The vertical distance from C to D is 3.

To find the coordinates of B, we will count 4 units right and 3 units up.

Vertex B is located at \left(2,-4\right).

Reflect and check

Completing the polygon, we can see that this does appear to be a rhombus.

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b

Verify ABCD is a rhombus.

Worked Solution
Create a strategy

To verify this is a rhombus, we need to prove all sides are the same length. We can do this using the distance formula d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}.

Apply the idea
\displaystyle AB\displaystyle =\displaystyle \sqrt{\left(-2-2\right)^2+\left(-1-\left(-4\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(-4\right)^2+\left(3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5
\displaystyle BC\displaystyle =\displaystyle \sqrt{\left(2-\left(-2\right)\right)^2+\left(-4-\left(-7\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(4\right)^2+\left(3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5
\displaystyle CD\displaystyle =\displaystyle \sqrt{\left(-2-\left(-6\right)\right)^2+\left(-7-\left(-4\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(4\right)^2+\left(-3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5
\displaystyle DA\displaystyle =\displaystyle \sqrt{\left(-6-\left(-2\right)\right)^2+\left(-4-\left(-1\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(-4\right)^2+\left(-3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5

Since \overline{AB}\cong \overline{BC} \cong \overline{CD} \cong \overline{DA}, the quadrilateral is a rhombus.

Reflect and check

If the question had asked us to verify that it was a parallelogram instead, we would have needed to find the slopes of each side to prove there are 2 sets of parallel sides.

Idea summary

If we know the coordinates of the vertices of a polygon, we can find the most precise classification using the lengths and slopes of line segments joining pairs of vertices.

  1. Find the slope of each side to determine if any sides are parallel or perpendicular.
  2. Find the length of each side to determine if any sides are congruent.

Rectangles, rhombuses, and squares

The following are special types of parallelograms, with specific properties about their sides, angles, and/or diagonals that help identify them:

Rectangle

A quadrilateral containing four right angles

A quadrilateral with four right angles.

These are three examples of rectangles:

Three rectangles with different orientations, lengths, and width.
Square

A quadrilateral with four right angles and four congruent sides

A quadrilateral with four congruent sides and four right angles.
Rhombus

A quadrilateral containing four congruent sides

A quadrilateral with four congruent sides.

These are two examples of rhombi:

Two rhombi with different orientations, lengths, and widths.

Exploration

Explore the applet by dragging the vertices of the polygons.

Loading interactive...
  1. Which polygon(s) are always rectangles? Can you create a rectangle with any of the polygons?
  2. How do you know which polygon(s) form a rhombus versus a square?
  3. Which polygon(s) are always parallelograms? Can you create a parallelogram with any of the polygons?

The following theorems relate to the special parallelograms:

Rectangle diagonals theorem

A parallelogram is a rectangle if and only if its diagonals are congruent

Rectangle A B C D with diagonals A C and B D. The four segments formed from each of the vertex to the point of intersection of the diagonals are all congruent.
Rhombus diagonals theorem

A parallelogram is a rhombus if and only if its diagonals are perpendicular

Rhombus P Q R S with diagonals P R and Q S drawn. Q S is perpendicular to P R.
Rhombus opposite angles theorem

A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles

Rhombus P Q R S with diagonals P R and Q S drawn. The diagonals bisect angles P, Q, R, and S.

Squares have the same properties as both a rectangle and rhombus.

A square with its diagonals drawn. The diagonals are perpendicular to each other. The four segments formed from each of the vertex to the point of intersection of the diagonals are all congruent.

Note that these theorems are for parallelograms, so if we are only told that a polygon is a quadrilateral, then they may not meet the conditions stated.

Examples

Example 5

List all classifications of quadrilaterals that apply to the figures. Explain your reasoning.

a
A quadrilateral with 4 right interior angles.
Worked Solution
Create a strategy

Recall what we know about parallelograms:

  • A quadrilateral is a parallelogram if and only if its opposite sides are congruent
  • A quadrilateral is a parallelogram if and only if its opposite angles are congruent
  • In a parallelogram, consecutive angles will be supplementary
  • A quadrilateral is a parallelogram if and only if its diagonals bisect each other

Recall the special types of parallelograms:

  • A rectangle is a quadrilateral with four right angles
  • A square is a quadtrilateral with four right angles and four congruent sides
  • A rhombus is a quadtrilateral containing four congruent sides
Apply the idea

The quadrilateral is a parallelogram and rectangle. A parallelogram's opposite angles are congruent, and a rectangle is a quadrilateral with four right angles by definition.

b
A quadrilateral with 4 congruent sides.
Worked Solution
Create a strategy

We are given information about the side lengths of the figure. Use this information to classify the figure.

Apply the idea

The figure is a parallelogram and rhombus. A parallelogram's opposite sides are congruent, and a quadrilateral with four congruent sides is a rhombus by definition.

c
A quadrilateral with 4 right interior angles and 4 congruent sides.
Worked Solution
Apply the idea

The figure can be classified as a parallelogram, square, rhombus, and rectangle because it has four right angles and consecutive congruent sides.

Reflect and check

Most specifically, since the angle measures are all right angles and the sides are all congruent, the figure is a square.

d
A quadrilateral with 2 pairs of adjacent congruent sides.
Worked Solution
Create a strategy

This figure cannot be classified as a parallelogram because its opposite sides are not congruent.

Apply the idea

The figure is a quadrilateral because it has four sides.

Example 6

Consider the diagram that illustrates the rhombus diagonals theorem: If a parallelogram is a rhombus, then its diagonals are perpendicular bisectors of one another.

A diagram showing 4 steps: step 1 shows rhombus A B C D; step 2 shows rhombus A B C D with diagonals A C and B D drawn, A C and B D intersect at point E, A E and E C are congruent as well as B E and E D; step 3 shows the same rhombus A B C D from step 2 with these additional information, angles D A C and A C D are congruent as well as angles A E D and D E C and angles A D E and E D C; step 4 shows the same rhombus A B C D from step 2 with these additional information, angles D A C and A C D are congruent as well as angles A D E and E D C, angles A E D and D E C are right angles.
a

Add reasoning to each step of the diagram.

Worked Solution
Create a strategy

Start with the given statement, and use the initial step to identify what is given. From there, the diagram provides more information to use when attempting to prove that the diagonals are perpendicular bisectors of one another.

Apply the idea
A figure titled Step 1 showing rhombus A B C D.

We are given a parallelogram that is a rhombus, where all sides are congruent and opposite sides are parallel to one another.

A figure titled Step 2 showing rhombus A B C D with diagonals A C and B D drawn. A C and B D intersect at point E. A E and E C are congruent as well as B E and E D.

The diagonals of the rhombus bisect one another since the diagonals of a parallelogram bisect one another.

A figure titled Step 3 showing rhombus A B C D with diagonals A C and B D drawn. A C and B D intersect at point E. A E and E C are congruent as well as B E and E D. Angles D A C and A C D are congruent as well as angles A E D and D E C and angles A D E and E D C.

The triangles formed by the diagonals are congruent by SSS congruence, so we know that their corresponding angles are also congruent by CPCTC.

A figure titled Step 4 showing rhombus A B C D with diagonals A C and B D drawn. A C and B D intersect at point E. A E and E C are congruent as well as B E and E D. Angles D A C and A C D are congruent as well as angles A D E and E D C. Angles A E D and D E C are right angles.

Since the triangles are congruent, we know that \angle AED \cong \angle CED and they are linear pairs along \overline{AC}. Since they are congruent and linear pairs, they must each be 90 \degree. This means that \overline{AC} \perp \overline{BD}, so the diagonals are perpendicular bisectors of one another.

b

Write a formal proof of the theorem.

Worked Solution
Create a strategy

Use the diagram from part (a), definitions, and theorems to write a formal proof.

Apply the idea

Consider the rhombus ABCD:

Rhombus A B C D with diagonals A C and B D drawn. The diagonals intersect at point E.
To prove: \overline{AC} is a perpendicular bisector of \overline{BD}
StatementsReasons
1.ABCD is a rhombusGiven
2.\overline{AC} is a bisector of \overline{BD}If a quadrilateral is a parallelogram, then its diagonals bisect each other.
3.\overline{AE} \cong \overline{EC} and \overline{BE} \cong \overline{ED}Definition of bisecting
4.\overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DA}Sides of rhombus are congruent
5.\triangle AED \cong \triangle AEB \cong \triangle CEB \cong \triangle CEDSide-side-side congruency theorem
6.\angle AED \cong \angle AEB \cong \angle CEB \cong \angle CEDCorresponding parts of congruent triangles theorem
7.m\angle AED = m\angle DECDefinition of congruence
8.m\angle AED + m\angle DEC=180 \degree\angle AED and \angle DEC are a linear pair
9.2 \cdot m\angle AED =180 \degreeSubstitution
10.m\angle AED =90 \degreeDivide by 2
11.m\angle AED = m\angle AEB = m\angle CEB = m\angle CED=90\degreeDefinition of congruence
12.\overline{AC} \perp \overline{BD}Definition of perpendicular
13.\overline{AC} and \overline{BD} are perpendicular bisectors of one anotherPerpendicular and bisect one another

Example 7

Prove that in the given rectangle ABCD, \overline{AC}\cong \overline{BD}:

Rectangle A B C D with diagonals A C and B D drawn.
Worked Solution
Create a strategy

It may be useful to mark up a diagram with information that we are given and that we already know using the definition of a rectangle, then use that to help build the proof.

Rectangle A B C D with diagonals A C and B D drawn. A B and D C are congruent as well as A D and B C. Angles A, B, C and D are right angles.
Apply the idea
A flow chart proof with 6 levels and the reason below each step. The third level has 2 steps. On the first level, the step is labeled A B C D, with reason Given. On the second level, the step is A B C D is parallelogram and angle A D C and angle B C D are right angles, with reason Definition of a rectangle. On the third level, the left step is labeled A D is congruent to B C, with reason If a quadrilateral is parallelogram, then its opposite sides are congruent, and the right step is labeled angle A D C is congruent to angle B C D, with reason All right angles are congruent. The left step on the thirs level proceeds to the fourth level while the right step goes directly to the fifth level. On the fourth level, the step is labeled D C is congruent to D C, with reason Reflexive property of congruence. On the fifth level, the step is labeled triangle A C D is congruent to triangle B D C, with reason Side angle side congruency theorem. On the sixth level, the step is labeled A C is congruent to B D, with reason C P C T C.

Example 8

Given:

  • ABCD is a rhombus
  • m\angle ACD=(5x+8)\degree
  • m\angle BCD=(12x+2)\degree

Complete the following:

Rhombus A B C D with diagonal A C drawn.
a

Solve for x.

Worked Solution
Create a strategy

Since parallelogram ABCD is a rhombus, the diagonals in a rhombus bisect a pair of opposite angles.

In other words, m\angle ACD = m\angle ACB and m\angle ACD+ m\angle ACB = m\angle BCD

We want to use this information to write an equation relating the two given angles.

\left(5x+8\right) + \left(5x+8\right)=\left(12x+2\right)

We want to use the above equation to solve for x.

Apply the idea
\displaystyle \left(5x+8\right) + \left(5x+8\right)\displaystyle =\displaystyle \left(12x+2\right)m\angle ACD+ m\angle ACB = m\angle BCD
\displaystyle 10x+16 \displaystyle =\displaystyle 12x+2Combine like terms
\displaystyle 10x+14\displaystyle =\displaystyle 12xSubtract 2 from both sides of equation
\displaystyle 14\displaystyle =\displaystyle 2xSubtract 10x from both sides of equation
\displaystyle 7\displaystyle =\displaystyle xDivide both sides of equation by 2
b

Solve for m\angle ABC.

Worked Solution
Create a strategy

\angle BCD and \angle ABC are consecutive angles.

Since ABCD is a rhombus and a rhombus is a parallelogram, we know that consecutive angles are supplementary.

So m\angle BCD + m\angle ABC = 180\degree.

Use the given information and the value for x we solved for in part (a) to find m\angle BCD. Then use that information and the above equation to solve for m\angle ABC.

Apply the idea

Substituting 7 for x, we get m\angle BCD=\left(12 \left(7 \right )+2 \right)\degree.

So m\angle BCD = 86 \degree. Now we want to use this angle measure and the fact that consecutive angles are supplementary to solve for m\angle ABC.

\displaystyle m\angle BCD + m\angle ABC\displaystyle =\displaystyle 180\degreeConsecutive angles in a parallelogram are supplementary
\displaystyle 86\degree + m\angle ABC\displaystyle =\displaystyle 180\degreeSubstitution
\displaystyle m\angle ABC\displaystyle =\displaystyle 94\degreeSubtract 86\degree from both sides of the equation

Example 9

If \overline{AB} \cong \overline{CD}, show that ABCD is not a rhombus.

Quadrilateral A B C D. A B has a length of 2 x plus 5, B C has a length of 3 x minus 2, C D has a length of 4 x minus 17, and D A has a length of 27.
Worked Solution
Create a strategy

We know that a rhombus has four congruent sides by definition, so to show that the figure is not a rhombus, we need to show that the side lengths are not congruent.

If \overline{AB} \cong \overline{CD}, we know that AB = CD by the definition of congruence. Start by solving the equation 2x+5=4x-17 for x, then use the value of x to determine the side lengths of the quadrilateral.

Apply the idea
\displaystyle 2x+5\displaystyle =\displaystyle 4x-17\overline{AB} = \overline{CD}
\displaystyle 2x+22\displaystyle =\displaystyle 4xAdd 17 to both sides of the equation
\displaystyle 22\displaystyle =\displaystyle 2xSubtract 2x from both sides of the equation
\displaystyle x\displaystyle =\displaystyle 11Divide both sides of the equation by 2

Since x=11, we know that BC=3x-2=3 \left(11 \right) -2= 22-2 = 31. Since AD \neq BC, ABCD is not a rhombus because we have shown that at least two of the sides are not congruent.

Idea summary

Use the theorems relating to the special parallelogrmas to solve problems:

  • Rectangle diagonals theorem: A parallelogram is a rectangle if and only if its diagonals are congruent
  • Rhombus diagonals theorem: A parallelogram is a rhombus if and only if its diagonals are perpendicular
  • Rhombus opposite angles theorem: A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles

Outcomes

G.CO.D.13

Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.

G.GPE.B.4

Use coordinates to prove simple geometric theorems algebraically.

G.GPE.B.5

Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems.

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