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STANDARDS OF LEARNING

CONTENT REVIEW NOTES

GEOMETRY

3rd Nine Weeks, 2018-2019

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OVERVIEW

Geometry Content Review Notes are designed by the High School Mathematics Steering Committee as a

resource for students and parents. Each nine weeks’ Standards of Learning (SOLs) have been identified and a

detailed explanation of the specific SOL is provided. Specific notes have also been included in this document

to assist students in understanding the concepts. Sample problems allow the students to see step-by-step models

for solving various types of problems. A “ ” section has also been developed to provide students with the

opportunity to solve similar problems and check their answers.

The document is a compilation of information found in the Virginia Department of Education (VDOE)

Curriculum Framework, Enhanced Scope and Sequence, and Released Test items. In addition to VDOE

information, Prentice Hall Textbook Series and resources have been used. Finally, information from various

websites is included. The websites are listed with the information as it appears in the document.

Supplemental online information can be accessed by scanning QR codes throughout the document. These will

take students to video tutorials and online resources. In addition, a self-assessment is available at the end of the

document to allow students to check their readiness for the nine-weeks test.

The Geometry Blueprint Summary Table is listed below as a snapshot of the reporting categories, the number of

questions per reporting category, and the corresponding SOLs.

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Polygons G.10 The student will solve problems, including practical problems, involving angles

of convex polygons. This will include determining the a) sum of the interior and/or exterior angles; b) measure of an interior and/or exterior angle; and c) number of sides of a regular polygon.

Polygons A convex polygon is defined as a polygon

with all its interior angles less than 180°.

This means that all the vertices of the polygon will point

outwards, away from the interior of the

shape.

A non-convex (concave) polygon is

defined as a polygon with one or more interior

angles greater than 180°. It looks like a vertex has been 'pushed in' towards the inside of the polygon.

A regular polygon is a polygon that is

equiangular (all angles are equal in measure)

and equilateral (all sides have the same

length).

A regular polygon is a polygon that is equiangular (all

angles are equal in measure) and

equilateral (all sides have the same

length).

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Interior and

Exterior Angles

n - # of sides

Sum of measures

of interior s

Each Interior Angle

Sum of

the Exterior Angles

Each Exterior Angle

REGULAR

POLYGONS

𝑆 = (𝑛 − 2) ∙ 180

(𝑛 − 2) ∙ 180

𝑛

3600 3600

𝑛

IRREGULAR POLYGONS 𝑆 = (𝑛 − 2) ∙ 180

Will vary based on the

algebraic or numerical

expressions

3600 Supplementary to

each of the

corresponding

interior angles

Example 1: Given a regular nonagon (9 sided convex polygon), what are the following measures?

a. The sum of the interior angles 𝑆 = (𝑛 − 2) ∙ 180

𝑆 = (9 − 2) ∙ 180

𝑆 = (7) ∙ 180

𝑆 = 1,260 b. Each interior angle

1,260

9= 1400

c. The sum of the exterior angle 3600

d. Each exterior angle 3600

9= 400

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Example 2: What are the values of x and y ?

a.

The interior angle, (5x + 5), and exterior angle, y, are supplementary. Therefore,

Example 3: Each interior angle of a regular polygon is 𝟏𝟑𝟓𝟎. How many sides does the polygon have?

𝐸𝑎𝑐ℎ 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒 =(𝑛 − 2) ∙ 180

𝑛

135 =(𝑛 − 2) ∙ 180

𝑛

135𝑛 = (𝑛 − 2) ∙ 180

135𝑛 = 180𝑛 − 360

−45𝑛 = −360

𝑛 = 8

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Polygons

1. What are the interior and exterior angle measures of a regular heptagon?

2. Given the 8-sided convex polygon, What is the value of n ?

3. Each interior angle of a regular polygon is 1500. How many sides does the polygon have?

Quadrilaterals G.9 The student will verify and use properties of quadrilaterals to solve problems,

including practical problems.

Properties of Quadrilaterals

Quadrilateral Properties

Parallelogram

Opposite Sides are Congruent

Consecutive Angles are Supplementary

Opposite Angles are Congruent

Diagonals Bisect Each Other

Rhombus

A parallelogram with 4 congruent sides

Diagonals are perpendicular

Each diagonal bisects opposite angles

Rectangle A parallelogram with 4 right

angles

Diagonals are congruent

Square A parallelogram with 4 congruent

sides and 4 right angles

Trapezoid

Exactly one pair of parallel sides

Midsegment is parallel to bases

Length of the midsegment is the average of the lengths of the bases

Isosceles Trapezoid

Legs are congruent

Base angles are congruent

Diagonals are congruent

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Example 1: ABCD is a parallelogram, solve for y.

Given: 𝐷𝑋̅̅ ̅̅ = 3𝑦 − 7 𝐵𝑋̅̅ ̅̅ = 8𝑦 − 27 Example 2: Based on the given information, can you prove that DEFG is a parallelogram? Example 3: Find the measure of the numbered angles in the rhombus.

∠1 = 90°

The diagonals of a rhombus are perpendicular.

∠1 + ∠4 + 32° = 180° 90° + ∠4 + 32° = 180°

∠4 = 58°

Triangle Angle-Sum Theorem

∠3 = ∠4 ∠4 = 58°

Alternate Interior Angles are Congruent

∠2 = 32° Diagonals of a rhombus bisect opposite angles.

Diagonals of a parallelogram bisect each other. Therefore 𝐷𝑋̅̅ ̅̅ = 𝐵𝑋̅̅ ̅̅

3𝑦 − 7 = 8𝑦 − 27

5𝑦 = 20

𝑦 = 4

You can show that 𝛥𝐷𝐸𝐺 ≅ 𝛥 𝐹𝐺𝐸 by Angle Side Angle.

Because corresponding parts of congruent triangles are congruent you

can show that

𝐷𝐸̅̅ ̅̅ ≅ 𝐺𝐹̅̅ ̅̅ 𝑎𝑛𝑑 𝑡ℎ𝑎𝑡 𝐷𝐺̅̅ ̅̅ ≅ 𝐸𝐹̅̅ ̅̅ . Once you show that both pairs of

opposite sides are congruent, you can say that DEFG is a parallelogram.

Scan this QR code to go to a video tutorial on Quadrilaterals.

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Quadrilaterals 1. What value of x will make the figure at the right a rectangle? 2. Janet is making a garden in the shape of a rhombus. One pair of opposite angles each measure 70°. What measure does each of the other opposite pair of angles measure? (Hint: Draw a picture.) It is often easier to classify geometric figures when they are drawn in the coordinate plane. Using slopes, distances and midpoints can help you with this.

Formula Example

Distance Formula

𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

Midpoint Formula

𝑀 = ( 𝑥1 + 𝑥2

2 ,

𝑦1 + 𝑦2

2 )

Find distance from A to B.

A (-2, -1) B (6, 3)

𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

𝑑 = √(6 − (−2))2 + (3 − (−1))2

𝑑 = √(8)2 + (4)2

𝑑 = √64 + 16

𝑑 = √80

Find the midpoint of AB.

A (-2, -1) B (6, 3)

𝑀 = ( 𝑥1 + 𝑥2

2 ,

𝑦1 + 𝑦2

2 )

𝑀 = ( (−2) + 6

2 ,

(−1) + 3

2 )

𝑀 = ( 4

2 ,

2

2 )

𝑀 = (2 , 1)

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Slope Formula

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1

Example 4: Is figure TRAP an isosceles trapezoid? Example 5: Is figure GRAM a square?

Find the slope of AB.

A (-2, -1) B (6, 3)

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1

𝑚 = 3 − (−1)

6 − (−2)

𝑚 = 4

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𝑚 = 1

2

In order to be an isosceles trapezoid, the legs must

be the same length. Therefore 𝑇𝑃̅̅̅̅ must equal 𝑅𝐴̅̅ ̅̅ .

Use the distance formula to determine if this is true.

Find distance from T to P.

T (-1, 3) P (-3, -2)

𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

𝑑 = √(−3 − (−1))2 + (−2 − 3)2

𝑑 = √(−2)2 + (−5)2

𝑑 = √4 + 25

𝑑 = √29

Find distance from R to A.

R (4, 3) A (5, -2)

𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

𝑑 = √(5 − 4)2 + (−2 − 3)2

𝑑 = √(1)2 + (−5)2

𝑑 = √1 + 25

𝑑 = √26

These distances are not the same therefore TRAP is NOT an isosceles trapezoid.

To be a square we must show that all sides are the same length, and that all sides meet at right angles

(are perpendicular to one another). Remember that for two sides to be perpendicular,

their slopes must be negative reciprocals.

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All of the sides meet at right angles because 1 and -1 are negative reciprocals of each other.

All of the sides will also have the same length (√50 ), therefore GRAM is a square.

Quadrilaterals 3. What figure is formed by the points (-1, -3), (1, 2), (7, 3) and (5, 5) 4. What figure is formed by connecting the midpoints of figure RECT?

Find the slope of AR.

A (1, -3) R (6, 2)

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1

𝑚 = 2 − (−3)

6 − 1

𝑚 = 5

5

𝑚 = 1

Find the slope of MA.

M (-4, 2) A (1, -3)

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1

𝑚 = −3 − 2

1 − (−4)

𝑚 = −5

5

𝑚 = −1

Find the slope of GR.

G (1, 7) R (6, 2)

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1

𝑚 = 2 − 7

6 − 1

𝑚 = −5

5

𝑚 = −1

Find the slope of GM.

G (1, 7) R (-4, 2)

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1

𝑚 = 2 − 7

−4 − 1

𝑚 = −5

−5

𝑚 = 1

Scan this QR code to go to a video tutorial on Coordinate

Geometry.

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Circles G.11 The student will solve problems, including practical problems, by applying

properties of circles. This will include determining a) angle measures formed by intersecting chords, secants, and/or tangents; b) lengths of segments formed by intersecting chords, secants, and/or tangents; c) arc length; and d) area of a sector.

The measure of a minor arc is equal to the measure of its corresponding central angle. You can add adjacent arc measures to find the measure of combined arc.

Example 1: What is the measure of 𝑫𝑪,̂ 𝑩𝑬,̂ 𝑩𝑫,̂ 𝑎𝑛𝑑 𝑫𝑬�̂� ?

You name a circle by its center. This is Circle X (ʘ X).

𝐴𝐵̅̅ ̅̅ is a diameter

𝑋𝐶̅̅ ̅̅ is a radius

𝐷𝐸̅̅ ̅̅ is a chord

∠𝐵𝑋𝐶 is a central angle (an angle whose vertex

is the center of a circle)

𝐵𝐶�̂� is a semicircle (an arc that is half of a circle)

𝐶�̂� is a minor arc (an arc that is less than a

semicircle)

𝐶𝐵�̂� is a major arc (an arc whose measure is

greater than 180° (a semicircle))

You name a minor arc by its endpoints.

You name a semicircle or major arc by its

endpoints and another point on the arc.

𝑩�̂� = 180° because it is a semicircle

𝐵�̂� + 𝑫�̂� = 180° therefore 𝑫�̂� = 𝟏𝟓𝟎°

𝐷�̂� + 𝑩�̂� = 180° therefore 𝑩�̂� = 𝟏𝟐𝟖°

𝑫𝑬�̂� = 𝐷�̂� + 𝐵�̂� + 𝐵�̂�

𝑫𝑬�̂� = 52° + 128° + 30°

𝑫𝑬�̂� = 𝟐𝟏𝟎°

You could have also found the measure of

𝐷𝐸�̂� by 180° + 𝐵�̂� (because 𝐷𝐸𝐵 = 180°̂ )

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The circumference of a circle is the measure of the distance around the outside of the circle. The formula for finding the circumference of a circle is

𝐶 = 2𝜋𝑟 𝑜𝑟 𝐶 = 𝜋𝑑 Use the circumference along with arc measure to find the length of a given arc.

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑟𝑐 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑟𝑐

360 ∙ 2𝜋𝑟

Example 2: What is the length of 𝐷�̂�, given 𝐷𝐵�̂� = 204° ?

Circles

𝐷�̂� = 360° − 𝐷𝐵�̂�

𝐷�̂� = 360° − 204°

𝐷�̂� = 156°

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐷�̂� = 156

360 ∙ 2 𝜋(3)

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐷�̂̂� = 2.6 𝜋 𝑓𝑡

1. Given that 𝑍𝑊̅̅ ̅̅ ̅ and 𝑉𝑌̅̅ ̅̅ are diameters of ʘ 𝐴, find the measures of all minor arcs of ʘ 𝐴.

2. Given that 𝑉𝑌̅̅ ̅̅ = 12 𝑖𝑛𝑐ℎ𝑒𝑠, find the length of 𝑋�̂�. Express in terms of 𝜋.

3. What is the circumference of ʘ 𝐵?

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The area of a circle can be found using the formula 𝐴 = 𝜋 𝑟2.

The sector of a circle is the region that is bounded by two radii. To find the area of the

sector of a circle use the formula 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑟𝑐

360 ∙ 𝜋 𝑟2.

Example 3: Find the area of sector BOC. Leave your answer in terms of 𝜋 .

Sometimes you will be asked to find the area of a segment of a circle. A segment is

made by joining the endpoints of an arc as shown in the picture below the shaded area

is the segment of the circle.

To find the area of the segment, use the radii from its endpoints to form a triangle.

To find the area of a sector we will use the formula

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑟𝑐

360 ∙ 𝜋 𝑟2

The measure of the arc is 90°, and the radius is 6 in.

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 𝐵𝑂𝐶 = 90

360 ∙ 𝜋 (6)2

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 𝐵𝑂𝐶 = 9 𝜋

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Example 4: Find the area of the shaded segment.

In the picture below 𝐴𝐵 ⃡ is tangent to ʘ 𝑂. This means that 𝐴𝐵 ⃡ is in the same plane as ʘ 𝑂 and intersects the circle in exactly one place. This place is point 𝑃, and is called the point of tangency. If a line is tangent to a circle, then that line is perpendicular to the radius of the circle.

Scan this QR code to go to a video tutorial on Areas of

Circles and Sectors.

Let’s start by finding the area of the sector that

includes the shaded segment.

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 = 90

360 ∙ 𝜋 (10)2

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 = 25𝜋 𝑖𝑛2

To find the area of the shaded region we need to

subtract the area of the triangle from the area

of the sector.

Area of a triangle is found by the formula

𝐴 = 1

2𝑏ℎ.

The base and height of this triangle are both 10.

𝐴 = 1

2(10)(10) = 50 𝑖𝑛2

The area of the shaded segment is the area of the sector with the area of the triangle subtracted.

𝐴𝑟𝑒𝑎 𝑠ℎ𝑎𝑑𝑒𝑑 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 25𝜋 − 50

𝐴𝑟𝑒𝑎 𝑠ℎ𝑎𝑑𝑒𝑑 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 ≈ 28.54 𝑖𝑛2

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Example 5: Is 𝑃𝑄̅̅ ̅̅ tangent to ʘ 𝐴 at 𝑄?

Circles 4. Find the area of the shaded region. Round to the nearest hundredth. 5. Find the radius of ʘ𝑀.

If 𝑃𝑄̅̅ ̅̅ is tangent to ʘ𝐴 at 𝑄 then ∆𝐴𝑄𝑃 must

be a right triangle.

Use the Pythagorean Theorem to determine

if ∆𝐴𝑄𝑃 is a right triangle.

Side 𝐴𝑃̅̅ ̅̅ is 8 + 9 = 17

82 + 152 ≟ 172

64 + 225 ≟ 289

289 = 289

Since ∆𝐴𝑃𝑄 is a right triangle, that means

that 𝑃𝑄̅̅ ̅̅ is tangent to ʘ𝐴 at 𝑄.

Scan this QR code to go to a

video tutorial on Tangent Lines.

°

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Theorems about Chords and Arcs

Within a circle, or in congruent circles, congruent central angles have

congruent arcs.

The converse is also true.

Within a circle or in congruent circles, congruent central angles have

congruent chords.

The converse is also true.

Within a circle or in congruent circles, congruent chords have congruent

arcs.

The converse is also true.

Within a circle or in congruent circles, chords equidistant from the center or

centers are congruent.

The converse is also true.

In a circle, if a diameter is perpendicular to a chord, then it

bisects the chord and its arc.

In a circle, if a diameter bisects a chord (that is not a diameter), then it

is perpendicular to the chord.

If ∠𝐷𝑂𝐸 ≅ ∠𝐺𝑂𝐹, then

𝐷�̂� ≅ 𝐺�̂�.

If 𝐷�̂� ≅ 𝐺�̂�, then

∠𝐷𝑂𝐸 ≅ ∠𝐺𝑂𝐹.

If ∠𝐷𝑂𝐸 ≅ ∠𝐺𝑂𝐹, then

𝐷𝐸̅̅ ̅̅ ≅ 𝐺𝐹̅̅ ̅̅ .

If 𝐷𝐸̅̅ ̅̅ ≅ 𝐺𝐹̅̅ ̅̅ , then

∠𝐷𝑂𝐸 ≅ ∠𝐺𝑂𝐹.

If 𝐷𝐸̅̅ ̅̅ ≅ 𝐺𝐹̅̅ ̅̅ , then 𝐷�̂� ≅

𝐺�̂�

If 𝐷�̂� ≅ 𝐺�̂�, then 𝐷𝐸̅̅ ̅̅ ≅

𝐺𝐹̅̅ ̅̅ .

If 𝑂𝑋̅̅ ̅̅ ≅ 𝑂𝑌̅̅ ̅̅ , then 𝐴𝐵̅̅ ̅̅ ≅

𝐶𝐷̅̅ ̅̅

If 𝐴𝐵̅̅ ̅̅ ≅ 𝐶𝐷̅̅ ̅̅ , then 𝑂𝑋̅̅ ̅̅ ≅

𝑂𝑌̅̅ ̅̅ .

If 𝑋𝑌̅̅ ̅̅ is a diameter and

𝑋𝑌̅̅ ̅̅ ┴ 𝐴𝐵̅̅ ̅̅

Then 𝐴𝐶̅̅ ̅̅ ≅ 𝐶𝐵̅̅ ̅̅ and

𝐴�̂� ≅ 𝐵�̂�.

If 𝑋𝑌̅̅ ̅̅ is a diameter and

𝐴𝐶̅̅ ̅̅ ≅ 𝐶𝐵̅̅ ̅̅ .

Then 𝑋𝑌̅̅ ̅̅ ┴ 𝐴𝐵̅̅ ̅̅ .

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In a circle, the perpendicular bisector of a chord contains the center of a

circle.

Example 6: Given ʘ𝐴 ≅ ʘ𝐷, and ∠𝐷𝐸𝐹 ≅ ∠𝐴𝐵𝐶. How can you show that ∠𝐴 ≅ ∠𝐷

Theorems about Angles and Segments

The measure of an inscribed angle is half the measure of its intercepted arc.

The measure of an angle formed by a tangent and a chord is half the measure

of the intercepted arc.

The measure of an angle formed by two lines that intersect inside a circle is half

the sum of the measure of the intercepted arcs.

Because the circles are congruent, you can

say that their radii are congruent. Because

the two congruent angles are across from

these radii, you can say that the other

angles across from the radii (∠𝐴𝐶𝐵 ≅ ∠𝐷𝐹𝐸)

are also congruent. If you subtracted the two

“known” angles from 180° you would have

the angle measure of the central angle.

These would have to be the same.

If 𝑋𝑌̅̅ ̅̅ is the

perpendicular bisector

of 𝐴𝐵̅̅ ̅̅

Then 𝑋𝑌̅̅ ̅̅ contains the

center of the circle.

𝑚∠𝑍 = 1

2𝑚 𝑋�̂�

𝑚∠𝐶 = 1

2𝑚 𝐴�̂�

𝑚∠1 = 1

2 (𝑥 + 𝑦)

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The measure of an angle formed by two lines that intersect outside of a circle is

half the difference of the measures of the intercepted arcs.

For a given point and circle, the product of the lengths of the two segments from the point to the circle is constant along

any line through the point and circle

Case I Case II Case III (𝑒 + 𝑓)𝑓 = (𝑔 + ℎ)ℎ

𝑎 ∙ 𝑏 = 𝑐 ∙ 𝑑 (𝑦 + 𝑥)𝑦 = 𝑧2

Example 7: Find the value of each variable. Example 8: Find the value of x.

𝑚∠1 = 1

2 (𝑥 − 𝑦)

𝑚∠𝑎 =1

2(70°) = 35°

𝑚∠𝑏 =1

2(150°) = 75°

Angle c is a vertical angle with the third angle in the

triangle that includes ∠’s a and b.

𝑚∠𝑐 = 180° − 35° − 75° = 70°

2(𝑥 + 2) = 62

2𝑥 + 4 = 36

2𝑥 = 32

𝑥 = 16

Scan this QR code to go to a video tutorial on Angle Measures

and Segment Lengths.

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Circles 6. What is the 𝑚∠𝑋 ? 7. What is the value of x?

°

°

°

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Answers to the problems: Polygons

1. 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 ≈ 128.60, 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 ≈ 51.40

2. n = 140°

3. 12 𝑠𝑖𝑑𝑒𝑠

Quadrilaterals

1. 𝑥 = 16

2. 1100

3. trapezoid

4. rhombus

Circles

1. 𝑉�̂� = 41°, 𝑋�̂� = 54°, 𝑉�̂� = 139° 2. 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑋�̂� = 1.8 𝜋

3. 𝐶 = √61 𝜋

4. ≈ 128.28 𝑖𝑛2 5. 𝑟𝑎𝑑𝑖𝑢𝑠 − 𝑥 = 3 6. 𝑥 = 100° 7. 𝑥 = 15