A LOOK at Geo - Hanlon Math · 2017-01-19 · altitude is geo mean 16 (CX)2 = 4(16) (CX)2 = 64 CX =...

A LOOK at Geometry; Review Packet

Lines, line segments

Segment Addition Postulate (sum of the parts equal the whole) Given segment AC, if B lies on AC and is between A and C, the AB + BC = AC.

Example 1 In the diagrams on the left above, find the value of x.

Using the Segment Addition Postulate, AB + BC = AC 8 + x = 12 x = 4

Example 2 In the diagram on the right above, find DE.

Using the Segment Addition Postulate, DE + EF = DF (2x+3) + (x+2) = 20 3x + 5 = 20 3x = 15 x = 5 Since x = 5, DE = 2x + 3, DE = 13

Midpoint Given a line segment AC, point M is the midpoint if and only if AM = MC

Example 3 In the diagram on the left above, M is the midpoint, find AM. If the AC = 24, then AM = 12.

Copyright © 2003 Hanlonmath Page of 1 20

-------------------20---------------------------

x + 22x + 3x

------------------12---------------------------

8

FEDCBA

------------------28---------------------

x + 4

----------------24------------------------

FMDM CA

Example 2 In the diagram above on the right, M is the midpoint of segment DF, DF = 28 and DM = x + 4, find the value of x.

Using the midpoint, DM = 1/2 DF. So DM =12 x + 4 = 12 x = 8

Angles Classification Acute angles whose measure is less than 90˚ Right angles measure 90˚ Obtuse angles are angles greater than 90˚ and less then 180˚ Straight Angle measures 180˚

Angle Addition Postulate (sum of the parts equal the whole) Given ∠AXC, if B is in the interior of the angle, then ∠AXB + ∠BXC = ∠AXC

Example 1 If m∠AXC = 80˚, find m∠AXB

Using the Angle Addition Postulate, 20˚ + ∠AXB = 80˚ ∠AXB = 60˚

Example 2 If m∠DYF = 65˚, find the m∠EYF

Using the Angle Addition Postulate, ∠DYE + ∠EYF = ∠DYF


(x – 5)˚

(2x + 10)˚20˚Y F

E

D

B

CX

A

(x–5) + (2x + 10) = 65 3x + 5 = 65 3x = 60 x = 20 Substituting, m∠EYF = 2x + 10; 50˚

Angle Pairs Complementary ∠’s – 2 ∠’s whose sum is 90˚ Supplementary ∠’s – 2∠ ’s whose sum is 180˚ Linear Pair – 2 ∠’s whose sum is 180˚

* If the exterior sides of two adj ∠’s lie in a line, they are supplementary * If the exterior sides of two adj ∠’s lie in perpendicular lines, they are complementary

Example 1 If ∠A and ∠B are complementary, find the m∠B if m∠A = 70˚

m∠A + m∠B = 90˚ 70˚ + m∠B = 90˚ m∠B = 20˚

Example 2 ∠P and ∠Q are a linear pair. ∠P = 5x + 30 and ∠Q = 2x + 10. Find the m∠P.

m∠P + m∠Q = 180˚ (5x+30) + (2x+10) = 180˚ 7x + 40 = 180 7x = 140 x = 20˚ Substituting, ∠P = 5x + 30, 130˚


* Vertical angles are congruent.

Example 3 In the figure on the left above, find m∠1

Since vertical angles are congruent, m∠1 = 60˚

Example 4 In the figure on the right above, find the value of x.

Since vertical angles are congruent; 8x = 3x + 10 5x = 10 x = 2

Angle Bisector – ! is said to be the bisector of ∠BAC if X lies in the interior and m∠BAX = m∠XAC

Example 5 m∠BAC = 70˚, find the value of x if m∠BAX = 3x + 5.

m∠BAX = 1/2 m∠BAC Substituting, 3x + 5 = 35˚ 3x = 30 x = 15

AX! "!!


Q

P

YB

A

(8x)˚

(3x + 10)˚

1 60˚

X

CA

B

Angles formed by Parallel Lines

If two parallel lines are cut by a transversal, then

The corresponding ∠’s are equal The alt int ∠’s are equal The alt ext ∠’s are equal The same side interior ∠’s = 180˚

Use ABBA to remember those angle relationships. All angles marked A are congruent; all angles marked B are congruent, and A + B = 180˚

In the next examples, use the diagram above, m || n

Example 1 If m∠2 = 70˚, find m∠6 ∠2 and ∠6 are corresponding ∠’s and are equal. m∠6 = 70˚


t

h

n

m

B

B

B

B

A

A

A

A

8765

4321

n

m

8765

4321

Example 2 If ∠3 = 10x – 15 and ∠6 =5x + 5, find the value of x.

∠3 and ∠6 are alternate interior ∠’s and are equal. 10x – 15 = 5x + 5 5x = 20 x = 4

Example 3 If ∠3 = 2x + 10 and ∠5 = 3x – 5 , find the value of x and the m ∠5.

∠3 and ∠5 are same side interior ∠’s, so their sum is 180˚ (2x + 10) + (3x – 5) = 180˚ 5x + 5 = 180˚ 5x = 175˚ x = 35˚ Substituting x = 35 in to 3x – 5; 100, so m∠5 = 100˚ Angles formed by Polygons

The sum of the interior angles of a triangle is 180˚ ∠1 + ∠2 + ∠3 = 180

The measure of each angle of an equiangular triangle have measure 60˚

The base ∠’s of an isosceles triangle are congruent The exterior ∠ of a triangle is equal to the sum of the 2 remote int ∠’s ∠3 = ∠1 + ∠2

If two angles of one triangle are equal to two angles of another triangle, the third angles must be equal.


3

2

1

3

2

1

The sum of the interior angles of a convex polygon is given by (n – 2)180˚

The sum of the exterior angles of a convex polygon is 360˚

Opposite ∠’s of a parallelogram are equal

Diagonals of a rhombus bisect a pair of opposite ∠’s

Example 1 In the triangle, if ∠1 = 5x, ∠2 = 2x + 10, and ∠3 = 3x –10, find the value of x and m∠2.

∠1 + ∠2 + ∠3 = 180 5x + (2x + 10) + (3x – 10) = 180˚ 10x = 180˚ x = 18˚ Substituting x = 18, 2x + 10; ∠2 = 46˚

Example 2 In the triangle, if m∠3 = 140˚. ∠2 = 3x – 5; ∠1 = 2x + 15. Find the value of x and m∠2

∠3 = ∠1 + ∠2 140 = (2x + 15) + (3x – 5) 140 = 5x + 10 130 = 5x 26 = x Substituting x = 26; 3x – 5; ∠2 = 47˚

Example 3 Find the measure of each angle of a regular octagon.

An octagon has 8 sides; (n – 2)180 is the sum of all the angles. (8 – 2)180 = 6(180) = 1080


3

2

1

3

2

1

Since there are 8 angles, divide by 8; there are 135˚ in each interior angle.

Example 4 If the interior angle of a regular polygon measures 150˚, how many sides does it have?

If the interior angles measures 150˚, then the exterior angle measures 30˚ (linear pair). Since the sum of the exterior angles is a constant (360˚), then 360 ÷ 30 = 13 equal angles - 12 sides.

Angles formed with Circles

Central ∠ is equal to its intercepted arc ∠1 = !

Inscribed ∠ is equal to HALF the intercepted arc

∠2 = !

If a quad is inscribed in a circle, the opposite ’s are supplementary

An inscribed ∠ in a semicircle is a right angle

Two secants intersect in the interior of a circle, the measure of the ∠ formed is equal to one half the sum of the intercepted arcs.

∠3 = !

AB!

12CD!

12(AB! +CD! )


B

A

1

D

C

2

D

C B

A

3

Two secants are drawn to a circle from an exterior point, the ∠ formed s equal to one half the difference of the intercepted arcs.

∠4 = !

Example 1 Find the measure of ∠A

∠A = ! = 40˚

Example 2 Find the m∠5 ∠5 = ! = 25˚

Example 3 Find ! , with the given information.

70˚ = !

140 = ! 110 = !

Example 4 Find the measure of ∠6

∠6 = ! = 40˚

12(CD! − AB! )

1280

12(40 +10)

AB!

12(AB! + 30)

(AB! + 30)

AB!

12(100 − 20)


D

CB

A

4

A80

1040 5

3070˚

B

A

620

100

Example 5 Find the value of x and the arc length DC.

20˚ = !

40 = ! 40 = 80 – 2x – 10 40 = 70 – 2x –30 = –2x 15 = x Substituting x = 15, arc DC = 2x + 10; 40˚

Congruence Methods

SSS HL SAS LL ASA HA AAS LA

Right Triangles - Similarity

If the altitude is drawn to the hypotenuse of a right triangle, the two triangles formed are similar to the given triangle and to each other.

∆ACB ~ ∆AXC and ∆ACB ~ ∆CXB, then ∆AXC ~ ∆CXB

12(80 − [2x +10])

(80 − [2x +10])


20˚

D

C

B

A

2x+1080

X

C

BA

Pythagorean Theorem – In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

3 Geometric Mean Theorems; ∆AXC ~ ∆CXB

If the altitude is drawn to the hypotenuse of a right triangle, it divides the hypotenuse into two segments, the altitude is the geometric mean between the lengths of the two segments.

If the altitude is drawn to the hypotenuse of a right triangle, the length of the leg of the right triangle is the geometric mean between the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.


X

C

BA

ABAC

= ACAX

and ABBC

= BCBX

X

C

BA

AXCX

= CXBX

Example 1 Find CX given AX = 4 and BX = 16

altitude is geo mean

(CX)2 = 4(16) (CX)2 = 64 CX = 8

Example 2 Find AC given AB = 20 and AX = 5

side is geo mean

(AC)2 = 5(20) (AC)2 = 100 AC = 10

Special Right Triangles

45-45-90 If each acute angle of a right triangle has measure 45˚, the hypotenuse is! times as long as the leg

30-60-90 If the acute angles of a right triangle have measures of 30˚ and 60˚; the hypotenuse is twice as long as the shorter leg and the longer leg is! times the shorter leg.

2

3


164 X

C

BA

4CX

= CX16

5-----------------------20--------------------------------

X

C

BA

20AC

= AC5

When working with special triangles, always find the short side first, then find the other sides.

The short side is opposite the smallest angle, the hypotenuse is opposite the hypotenuse, the long side is opposite the 60˚ angle.

Example 1 In a 45-45-90 triangle, find the hypotenuse if a side measures 5.

The hypotenuse is the ! times the leg; 5!

Example 2 Given a 30-60-90 triangle with side side opposite the 30˚ angle measures 4, find the length of the longer leg and the hypotenuse.

The short side is 4 is given The long side is 4! The hypotenuse is 8

Example 3 Given a 30-60-90 triangle, find the short side and the long side if the hypotenuse 10.

Find the short side first, the hypotenuse is 10, so the short side is 5 The long side is 5!

2 2

3

3


2x

x 3

x60˚

30˚

45˚

45˚

x 2

x

x

Right Triangle Trigonometry

SOHCAHTOA

sinA = ! cosA = ! tanA = !

Triangles

If a line is parallel to one side of a triangle and intersects the other 2 sides, it divides them proportionally.

The segment joining the midpoints of 2 sides of a triangle is parallel to the third side and half the length of the third side

M1M2 = !

If a ray bisects an angle of a triangle, it divides the opposite side into segments whose lengths are proportional to the lengths of the other two sides.

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

oppositehypotenuse

adjacenthypotenuse

oppositeadjacent

12AB


d

c

b

a

ab= cd

X

C

BAAXBX

= ACBC

M2M1

C

BA

Triangle Solutions - Law of Sines & Cosines

Law of Sines

Use the Law of Sines when you do not have a right triangle and you know: a. 2 ∠’s and a side or b. 2 sides and an ∠ opposite one of those sides

Law of Cosines c2 = a2 + b2 – 2ab cosC

Use the Law of Cosines when you do not have a right triangle and you know

a. 2 sides and included angle or b. 3 sides

Areas - Volumes

lw rectangle bh parallelogram ½ bh triangle ½(B+b)h trapezoid 1/2 ap regular polygon πr2 circle

To find the lateral area, find the area of all the sides.

To find surface area, find the lateral area and add the areas of the top and bottom

To find the volume of prisms, multiply the area of the base by the height of the prism.


aSinA = b

sinB = csinC

To find the volume of a pyramid, multiply the area of the base by the height of the pyramid, divide that result by 3.

Circumference is 2πr or d

Circles - Segments

When 2 chords intersect within a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other.

ab = cd

If 2 secants are drawn to a circle from an exterior point, the product of the lengths of one secant and its external segment is equal to the product of the lengths of the other secant and its external segment.

If a tangent and secant are drawn to a circle from an exterior point, the square of the length of the tangent is equal to product of the secant and its external segment.


dc

b

a

Y

X

C

B

A

BC ⋅BX = BA ⋅BY

Y

C

BA

(BC)2 = BA ⋅BY

Example 1 Find the value of x.

6x = 8 (3) 6x = 24 x = 4

Example 2 Find RX, if PX = 25 and MX = 20 and NX = 10


x

8

3

6

R

P

XN

M

MX ⋅NX = PX ⋅RX 20 ⋅10 = 25 ⋅RX 200 = 25 ⋅RX 8 = RX

Conic Sections In a quadratic equation, if only one variable is squared, the graph is a parabola.

if two variables are squared and the coefficients are the same, then it is a circle.

if two variables are squared and the coefficients are not equal but have the same sign, then it is an ellipse.

if two variables are squared and the coefficients have opposite signs, then it is a hyperbola

Example 1 Name the type of conic section based upon the coefficients of the quadratic term(s).

A) 5x2+ 5y2 + 10y – 30 = 0 B) 4x2 – 3y2 +10 = 0 C) x2 + 4y + 12 = 0 D) 3x2 + 5y2 + 6x – 12 = 0

A) The coefficients of the quadratic terms are equal circle B) The coefficients of the quadratic terms have opposite signs hyperbola C) There is only one quadratic term parabola D) The coefficients of the quadratic terms have the same sign and are not

equal ellipse

Example 2 Find the center and radius of a circle and sketch the curve of a circle x2 + y2 – 4x + 6y – 12 = 0

We know this is a circle because it was given to us and the coefficients of the quadratic terms are the same. To determine the center and radius, I have to write the equation in (x–h)2 + (y –k)2 = r2. To accomplish that, I will need to complete the squares.


Rewriting the equation, x2 – 4x + __ + y2 +6y + __ = 12 x2 – 4x + 4 + y2 +6y + 9 = 12 + 4 + 9 (x – 2)2 + (y + 3)2 = 52

This is a circle with center (2, –3) and radius 5.

Example 3 Find an equation of a curve whose vertex is (1, 0), focus (3,0), find the equation of the directrix and sketch the curve.

We know this will be a parabola because of the information given, focus and directrix. Because of the location of the vertex and focus, we also know this parabola will open to the right. So the y-term will be squared.

The equation will take the form, x – h = !

Substituting the coordinates of the vertex (1,0) and determining c = 2, the distance from the vertex to the focus gives us

Simplifying x – 1 = !

The equation of the directrix has to be two over from the vertex, so x = –1 is the equation of the directrix

Example 4 Find the center, foci, vertices and sketch the curve

!

14c(y − k)2

18y2

(x − 4)2

12+ (y − 6)

2

3= 3


2

–2

–4

–6

–8

5

(2, –3)

The right side of the equation must be equal to 1, so divide both sides by 3. Its an ellipse because of the “+” sign.

From this equation, we see that a = 6 and b = 3. We also know that b2 = a2 – c2

Substituting, 32 = 62 – c2

c2 = or c =

Now that we know the values of a, b and c, let’s do some arithmetic to find the vertices. The center, taken right from the original equation is (4, 6). The foci are located on the major axis, since 6 > 3, that would be the x-axis. The foci are located ! from the center (4, 6) on the x-axis. So the foci are (4+! , 6) and (4–! , 6).

The vertices are located 6 units and 3 units from the center on the major and minor axes, respectively. The vertices are (4+6, 6), (4 – 6, 6), (4, 6+3), and (4, 6–3). Simplifying those ordered pairs; (10,6), (–2,6), (4, 9), and (4, 3).

All I did to find all those coordinates was add or subtract. Now on to the graph.

I graph the center and the vertices, then I’m done!

If you know the formulas and can visualize your graph, answering these questions are actually very easy because the computations are mostly arithmetic as we could see from the last example.

27 3 3

3 33 3 3 3


10

8

6

4

2

5 5 10

(4, 6)(x− 4)26 + (y− 6)

2

3 =1

A LOOK at Geo - Hanlon Math · 2017-01-19 · altitude is geo mean 16 (CX)2 = 4(16) (CX)2 = 64 CX =...

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Transcript of A LOOK at Geo - Hanlon Math · 2017-01-19 · altitude is geo mean 16 (CX)2 = 4(16) (CX)2 = 64 CX =...