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Difficulty Level: At Grade Created by: CK-12

## Angles in Polygons

All Those Polygons - Although they have probably been taught it before, not all students will remember the names of the different polygons. There are not very many opportunities in life to use the word heptagon. Add these words to their vocabulary list.

This is most likely the first time they have been introduced to a polygon with a variable number of sides, an \begin{align*}n-\end{align*}gon. This notation can be used when referring to a polygon that does not have a special name in common use, like a 19-gon. It can also be used when the number of sides of the polygon is unknown.

Clockwise or Counterclockwise But Not Both - At this point in the class, student are usually good at recognizing vertical angles. They will understand that the exterior angles made by extending the sides of the polygon in a clockwise rotation are congruent, at each vertex, to the exterior angle formed by extending the sides counterclockwise. What they will sometimes do is include both of these angles when using the Exterior Angle Sum theorem. Reinforce that the number of exterior angles is the same as the number of interior angles and sides, one at each vertex.

Interior or Exterior - Interior and exterior angles come in linear pairs. If one of these angles is known at a particular vertex, it is simple to find the other. When finding missing angles in a polygon, students need to decide from the beginning if they are going to use the interior or exterior sum. Most likely, if the majority of the known angle measures are from interior angles they will use the interior sum. They need to convert the exterior angle measures to interior angle measures before including them in the sum. If there are more exterior angle measures given, they can convert the interior angle measures and use the sum of \begin{align*}360^\circ\end{align*}. It is important that they make a clear choice. They may mix the two types of angles in one summation if they are not careful.

Do a Double Check - Students often do not take the time to think about their answers. Going over the arithmetic and logic is one way to check work, but it is common to not recognize the error the second time either. A better strategy is to use other relationships to do the checking. In this lesson if the exterior sum was used, the work can be checked with the interior sum.

What’s the Interior Sum of a Nonagon Again? - If students do not remember the interior sum for a specific polygon, and do not remember the formula, they can always convert to the exterior angle measures using the linear pair relationship. The sum of the exterior angles is always \begin{align*}360^\circ\end{align*}. This strategy will work just as well as using the interior sum. Remind the students to be creative. When taking a test, they may not know an answer directly, but many times they can figure out the answer in an alternative way.

It is important to practice lots of problems like the following examples and to help student devise strategies to solve them.

Example 1: What is the measure of each interior angle of a regular \begin{align*}n-\end{align*}gon if the sum of the interior angles is \begin{align*}1080^\circ\end{align*}?

Answer: \begin{align*}135^\circ\end{align*} First the number of sides needs to be found, so set the sum equal to \begin{align*}(n-2)180^\circ\end{align*} and solve to find \begin{align*}n = 8\end{align*}. Now the total of \begin{align*}1080^\circ\end{align*} needs to be divided into 8 congruent angles: \begin{align*}\frac{1080^\circ}{8} = 135^\circ\end{align*}.

Example 2: If three angles in a pentagon have measures \begin{align*}115^\circ, 100^\circ\end{align*} and \begin{align*}125^\circ\end{align*} what are the measures of the two remaining angles if they are congruent to each other?

Answer: \begin{align*}100^\circ\end{align*} First we must determine the sum of the 5 angles in a pentagon: \begin{align*}(5-2)180^\circ = 540^\circ\end{align*}. Next, subtract the three known angles from the sum: \begin{align*}540^\circ -115^\circ -100^\circ -125^\circ = 200^\circ\end{align*}. Finally, divide the remaining measure by 2 to get \begin{align*}100^\circ\end{align*}.

Example 3: If the measure of one interior angle in a regular polygon is \begin{align*}168^\circ\end{align*}, how many sides does it have?

Answer: 30 The easiest way to solve this problem is to use the exterior angle sum. If the interior angle measure is \begin{align*}168^\circ\end{align*}, then the measure of each exterior angle is the supplement of this, or \begin{align*}12^\circ\end{align*}. Now simply divide the exterior sum, \begin{align*}360^\circ\end{align*} by \begin{align*}12^\circ\end{align*} to get 30, the number of sides.

Sketchpad Alternatives - Many students become particularly engaged in a topic when they are able to investigate it while playing around with the computer. Here are a couple of ways to use Geometers’ Sketchpad in the classroom as an alternative or supplement to direct instruction.

Angle Sum Conjecture – Have student make different convex polygons and measure the sum of their interior angles.

1. The students should observe that for each type of polygon, no matter how many were drawn, they all have the same interior angle sum. They also like to see that if they change the shape of the convex polygon (by dragging a vertex- making sure it stays convex) that the sum remains the same.
2. The students should drag a vertex of each polygon toward the center to create a concave polygon, and notice if the sum stays the same. (It won’t.)
3. Put the sums in order on the board: \begin{align*}180^\circ,360^\circ,540^\circ,\ldots\end{align*} Ask the students to find the pattern in this sequence of numbers. Lead them to discovering the Angle Sum formula, \begin{align*}(n-2)180^\circ\end{align*} from the pattern.

Exterior Angle Sum Conjecture – Have student make different convex polygons and measure the sum of their exterior angles. Using Sketchpad to extend the sides of the polygon helps students gain an understanding of where the exterior angle is in relation to the polygon.

1. Students should observe that the sum is always \begin{align*}360^\circ\end{align*}, regardless of the number of sides. Doing this for polygons with different numbers of sides will help them see why the exterior angle sum remains the same.
2. Students can again drag the vertex of a polygon and see that although the angles may change, the sum does not.

## Properties of Parallelograms

Tree Diagram - Most students will need practice working with the classification of quadrilaterals before they completely understand and remember all of the relationships. The Venn diagram is an important mathematical tool and should definitely be used to display the relationships among the different types of quadrilaterals. A tree diagram will also make an informative visual. Using both methods will reinforce the students’ understanding of quadrilaterals, and their ability to make good diagrams.

Parallel Line Properties - In Chapter Three: Parallel and Perpendicular Lines, the students learn about the relationships between the measures of the angles formed by parallel lines and a transversal. Many of the quadrilaterals studied in this section have parallel sides. The students can apply what they learned in chapter three to the quadrilaterals in this chapter. They may have trouble seeing the relationships because instead of lines the quadrilaterals are made of segments. Recommend that the students draw the figures on their papers and extend the sides of the quadrilaterals so they can see all four angles made by the intersection of the lines. These angles will be useful when looking for specific information about the quadrilateral.

Show Clear, Organized Work - When using the distance or slope formula to verify information about a quadrilateral on the coordinate plane, students will often do messy scratch work as if they are the only ones that will need to read it. In this situation, the work is a major part of the answer. They need to communicate their thoughts on the situation. They should write as if they are trying to convince the reader that they are correct. As students progress in their study of mathematics, this is more often the case than the need for a single numerical answer. They should start developing good habits now.

Symmetry - Most students have already studied symmetry at some point in their education. A review here may be in order. When studying quadrilaterals, symmetry is a good property to consider. Symmetry is also important when discussing the graphs of key functions that the students will be studying in the next few years. It will serve the students well to be adept in recognizing different types of symmetry.

Proofs Using Congruent Triangles - The majority of the proofs in this section use congruent triangles. The quadrilateral of interest is somehow divided into triangles that can be proved congruent with the theorems and postulates of the previous chapters. Once the triangles are known to be congruent, the definition of congruent triangles ensures that certain parts of the quadrilateral are also congruent. Students should be made aware of this pattern if they are having difficulty writing or understanding the proofs of the properties of various quadrilaterals. If they are still struggling they should spend some time reviewing Chapter Four: Triangles and Congruence.

The Diagonals of Parallelograms - The properties concerning the sides and angles of parallelograms are fairly intuitive, and students pick them up quickly. More emphasis should be placed on what is known, and not known about the diagonals. Students frequently try to use the incorrect fact that the diagonally of a parallelogram are congruent. Rectangles are the focus of an upcoming lesson, but demonstrating to students that the diagonals of a quadrilateral are only congruent in the special case where all the angles of the parallelogram are congruent. For a general parallelogram, the diagonals bisect each other. This can be shown nicely using a sketch in Geometer’s Sketchpad.

Proof Practice - The proofs in this section may seem a bit repetitive, but students will benefit from practicing these proofs since they review important concepts learned earlier in the course. To avoid loosing the students’ attention, find different ways of presenting the proofs. One idea is to divide the students into groups, and have each group demonstrate a different proof to the class.

Parallel or Congruent - When looking at a marked figure students will sometimes see the arrows that designate parallel segments and take that the segments to be congruent. This could be due to the misreading of the marks, or mistakenly thinking parallel always implies congruence. Warn students not to make this error. The last method of proof in this section which utilizes that one pair of sides are both congruent and parallel, along with an example of a trapezoid where the parallel sides are not congruent, will help students remember the difference.

Below are some additional examples to be shown in class.

Example 1: Quadrilateral \begin{align*}ABCD\end{align*} is a parallelogram. \begin{align*}AB=2x+5, \ BC=x-3\end{align*} and \begin{align*}DC=3x-10\end{align*}. Find the measures of all four sides of the quadrilateral.

Answer: \begin{align*}AB=CD=35\end{align*} and \begin{align*}BC=AD=12\end{align*}.

Encourage students to draw and label a diagram for problems such as these. Remind them that the name of a polygon lists the vertices in a circular order. This will help ensure that they match up correct pairs of congruent sides. Set the congruent sides equal to each other and solve for \begin{align*}x\end{align*}.

\begin{align*}2x+5&=3x-10\\ x&=15\end{align*}

Example 2: \begin{align*}JACK\end{align*} is a parallelogram. \begin{align*}m \angle A=(10x-60)^\circ\end{align*} and \begin{align*}m \angle C=(2x+45)^\circ\end{align*}. Find the measures of all four angles.

Answer: \begin{align*}m \angle C = m \angle J=77.5^\circ\end{align*} and \begin{align*}m \angle A = m \angle K = 102.5^\circ\end{align*}

Again, encourage students to sketch and label a diagram so they can see that \begin{align*}\angle A\end{align*} and \begin{align*}\angle C\end{align*} are consecutive angles in the parallelogram and therefore supplementary. Now, they can set the sum of the expressions for the measures of these two angles equal to \begin{align*}180^\circ\end{align*} and solve for \begin{align*}x\end{align*}.

\begin{align*}10x-60+2x+45&=180^\circ\\ x&=16 \frac{1}{4}\end{align*}

Example 3: \begin{align*}KATE\end{align*} is a parallelogram with a perimeter of 40 cm. \begin{align*}KA = 3x+8\end{align*} and \begin{align*}AT=x+4\end{align*}. Find the length of each side.

Answer: \begin{align*}KA=ET=14 \ cm\end{align*} and \begin{align*}AT = KE = 6 \ cm\end{align*}

After sketching a diagram, students should recognize that \begin{align*}KA\end{align*} and \begin{align*}AT\end{align*} are adjacent sides and therefore not necessarily congruent. Remind them that they were given the perimeter and help them figure out that they need to add two times each of the given side length expressions and set this sum equal to the perimeter of the parallelogram as shown below.

\begin{align*}2(3x+8)+2(x+4)&=40\\ x&=2\end{align*}

Example 4: \begin{align*}SAMY\end{align*} is a parallelogram with diagonals intersecting at point \begin{align*}X\end{align*}. \begin{align*}SX=x+5, XM=2x-7\end{align*} and \begin{align*}AX=12x\end{align*}. Find the length of each diagonal.

Answer: \begin{align*}SM=34 \ cm\end{align*} and \begin{align*}AY=288 \ cm\end{align*}

Students should use the fact that diagonals bisect each other in a parallelogram to solve this one. By setting the two parts of diagonal \begin{align*}SM\end{align*} each to each other they can find \begin{align*}x\end{align*} and then solve for the lengths of the diagonals.

\begin{align*}x+5&=2x-7\\ x&=12\end{align*}

Example 5: \begin{align*}JEDI\end{align*} is a parallelogram. \begin{align*}m \angle J = 2x + 60\end{align*} and \begin{align*}m \angle D = 3x + 45\end{align*}. Find the measures of the four angles of the parallelogram. Does this parallelogram have a more specific categorization?

Answer: All four angles measure \begin{align*}90^\circ\end{align*}. \begin{align*}JEDI\end{align*} is a rectangle.

The angles given here are opposite angles. By setting them equal we can solve for \begin{align*}x\end{align*} and find the angle measures.

\begin{align*}2x+60&=3x+45\\ x&=15\end{align*}

## Rectangles, Rhombuses and Squares

The Power of the Square - Students should know by the classification of quadrilaterals that all the theorems for parallelograms, rectangles, and rhombuses, also apply to squares. It is a good idea to talk about this in class though in case they have not put it together on their own. These theorems and the definition of a square can be combined to from some interesting exercises.

A Venn diagram can be used to show the relationships between the figures and their properties. This is also an excellent opportunity to review some logic and practice making conditional statements and checking their validity based on the Venn diagram. For example: If the quadrilateral is a square, then it is a parallelogram. This statement is true, but its converse is false.

Information Overload - Quite a few theorems are presented in this chapter. Remembering them all and which quadrilaterals they apply to can be a challenge for students. If they are unsure, and cannot check reference material, a test case can be drawn. For example: Do the diagonals of a parallelogram bisect the interior angles of that parallelogram? First they need to draw a parallelogram that clearly does not fit into any subcategory. It should be long and skinny, so no rhombi properties are mistakenly attributed to it. It should also be well slanted over, so as not to be mistaken for a rectangle. Now they can draw in the diagonals. It will be obvious that the diagonals are not bisecting the interior angles. They could also try to recreate the proof, but that will probably be more time consuming and it requires a bit of skill.

Also, students need to be reminded that they must show that the quadrilateral is a parallelogram before it can be determined that it is one of the special parallelograms. For example, just because the diagonals of the quadrilateral are congruent, it is not sufficient information to conclude the quadrilateral is a rectangle- it could be an isosceles trapezoid. If students can also show that one of the properties of parallelograms is true for the quadrilateral as well, such as opposite sides are parallel, then they can correctly conclude that the quadrilateral is a rectangle.

Example 1: \begin{align*}SQUR\end{align*} is a square and \begin{align*}X\end{align*} is the point where the diagonals meet. \begin{align*}QX=3x-9\end{align*} and \begin{align*}SX=2x\end{align*}. Find the length of both diagonals.

Answer: \begin{align*}SU=QR=36\end{align*}.

Since the diagonals in a square are equal and bisect each other, we can set \begin{align*}3x-9=2x\end{align*} and solve to get \begin{align*}x=9\end{align*}. Then \begin{align*}QX=18\end{align*} and both diagonals are 36.

Example 2: \begin{align*}DAVE\end{align*} is a rhombus with diagonals that intersect at point \begin{align*}X\end{align*}. \begin{align*}DX=3 \ cm\end{align*} and \begin{align*}AX=4 \ cm\end{align*}. How long is each side of the rhombus?

Answer: \begin{align*}DA = AV = VE = ED = 5 \ cm\end{align*}

The diagonals of a rhombus are perpendicular bisectors of each other so \begin{align*}DX\end{align*} and \begin{align*}AX\end{align*} are the legs of a right triangle. The hypotenuse of this right triangle is a side of the rhombus and is 5 cm by Pythagorean Theorem.

## Trapezoids and Kites

Average for the Median - Students who have trouble memorizing formulas may be intimidated by the formula for the length of the median of a trapezoid. Inform them that they already know this formula; it is just the average. The application of the formula makes since, the location of the median is directly between the two bases, and the length of the median is exactly between the lengths of the bases. They will have no problem finding values involving the median.

Where Are We? - It is easy for students to forget how what they are learning today relates to the chapter and to the class. Use the Venn diagram of the classification of quadrilaterals to orient them in the chapter. They are no longer learning about parallelograms, but have moved over to the separate trapezoid area. When student are able to organize their new knowledge, they are better able to retain and apply it.

Does it have to be Isosceles? - Students may have trouble remembering which theorems in this section apply only to isosceles trapezoids. Note that base angle, and diagonal congruence apply only to isosceles trapezoids, but the relationship of the length of the median to the bases is the same for all trapezoids.

Example 1: \begin{align*}TRAP\end{align*} is a trapezoid. The median has length 4 cm, and one of the bases has length 7 cm. What is the length of the other base?

There are two ways to approach this problem. One is to compare differences – seven is three more than four, so the other base must be three less than four. The second way is to set up and solve an equation as shown below.

\begin{align*}4 &= \frac{7+x}{2}\\ x &= 1\end{align*}

Example 2: \begin{align*}WXYZ\end{align*} is a trapezoid. The length of one base is twice the length of the other base, and the median is 9 cm. How long is each base?

Answer: The bases are 6 cm and 12 cm.

This problem can be solved by allowing one base to be \begin{align*}x\end{align*} and the other base to be \begin{align*}2x\end{align*}. Then solve the following equation:

\begin{align*}\frac{x+2x}{2} & = 9\\ x & =6\end{align*}

Only One Congruent Set - It is important to note that in a kite, only one set of interior angles are congruent, and only one of the diagonals is bisected. Sometimes students struggle with identifying where these properties hold. It is the nonvertex angles that are congruent, and the diagonal connecting the nonvertex angles that is bisected. The single line of symmetry of a kite shows both these relationships. Remind students to think of this line of symmetry and what it tells them about congruent pairs of angles and segments.

Break it Up - When working with a kite, it is sometimes easier to think of it as two isosceles triangles, or four right triangles, instead of one quadrilateral.

At this point in the class, students have had extensive experience working with isosceles triangles, and can easily apply the Base Angle theorem to see that the nonvertex angles of the kite are congruent. They have also seen that the diagonal segment between the vertex angles creates many symmetries in the triangle. If students think of the symmetries in the triangle, it will make sense to them that the vertex angles are bisected and that the diagonal connecting the nonvertex angles is bisected.

They can also think of a kite as four right triangles. This will help them remember that the diagonals are perpendicular, and remind them that the Pythagorean Theorem can be used to find missing segment measures. Noticing that the right triangles are in two congruent sets will help them identify congruent segments and angles.

Example 1: \begin{align*}KITE\end{align*} is a kite with \begin{align*}\overline{KE} \cong \overline{KI}\end{align*}. Prove that \begin{align*}\Delta KSE \cong \Delta KSI\end{align*}.

Statement Reason
\begin{align*}\overline{KE} \cong \overline{KI}\end{align*} Given
\begin{align*}\overline{ES} \cong \overline{SI}\end{align*} Kite Diagonal Theorem
\begin{align*}\angle KSE\end{align*} and \begin{align*}\angle KSI\end{align*} are right angles Definition of Perpendicular
\begin{align*}\angle KSE \cong \angle KSI\end{align*} Right Angle Theorem
\begin{align*}\Delta KSE \cong \Delta KSI\end{align*} HL

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