- 98 Views
- Uploaded on
- Presentation posted in: General

Geometry

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Constructions

Given:

A

B

Construct a segment congruent to a given segment.

This is our compass.

Procedure:

1. Use a straightedge to draw a line. Call it l.

Construct: XY = AB

Don’t change your radius!

2. Choose any point on l and label it X.

3. Set your compass for radius AB and make a mark on the line where B lies. Then, move your compass to line l and set your pointer on X. Make a mark on the line and label it Y.

l

X

Y

A

1) Draw a ray. Label it RY.

2) Using B as center and any radius, draw an arc intersecting BA and BC. Label the points of intersection D and E.

B

C

3) Using R as center and the SAME RADIUS as in Step 2, draw an arc intersecting RY. Label point E2 the point where the arc intersects RY

R

Y

Construct an angle congruent to a given angle

Given:

Procedure:

D

E

Construct:

D2

4) Measure the arc from D to E.

E2

5) Move the pointer toE2 and make an arc that that intersects the blue arc to get point D2

6) Draw a ray from R through D2

A

B

C

Using B as center and any radius, draw

and arc that intersects BA at X and BC at point Y.

3. Draw BZ.

How do I construct a

Bisector of a given angle?

Z

Given:

X

Y

Procedure:

2. Using X as center and a suitable radius, draw and arc.

Using Y as center and the same radius, draw an arc

that intersects the arc with center X at point Z.

B

A

2. Draw XY

How do I construct a perpendicular

bisector to a given segment?

X

Given:

Y

Procedure:

Using any radius greater than 1/2 AB, draw four arcs of

equal radii, two with center A and two with center B.

Label the points of intersection X and Y.

C

k

3. Draw CZ.

How do I construct a perpendicular bisector

to a given segment at a given point?

Z

Given:

X

Y

Procedure:

Using C as center and any radius, draw arcs intersecting

k at X and Y.

Using X as center and any radius greater than CX,draw an arc. Using Y as center and the same radius,

draw and arc intersecting the arc with center X at Z.

P

k

3. Draw PZ.

How do I construct a perpendicular bisector to a

given segment at a given point outside the line?

Given:

X

Y

Z

Procedure:

Using P as center, draw two arcs of equal radii that

intersect k at points X and Y.

Using X and Y as centers and a suitable radius, draw arcs

that intersect at a point Z.

P

k

Let A and B be two points on line k. Draw PA.

How do I construct a line parallel to a

given line through a given point?

1

l

Given:

A

B

Procedure:

At P, construct <1 so that <1 and <PAB are congruent

corresponding angles. Let l be the line containing the

ray you just constructed.

If the lines are Concurrent then they all intersect at the same point.

The point of intersection is called the“point of concurrency”

- Perpendicular bisectors
- Angle bisectors
- Altitudes
- Medians

- Circumcenter
- Incenter
- Orthocenter
- Centriod

Concurrent Lines

Point of Concurrency

Circumcenter

SP

Orthocenter

SP

Incenter

SP

Centroid

SP

1) Draw Ray OA

2) Construct a perpendicular through OA at point A.

3) Draw tangent line XY

Construct arcs 3 & 4 using point Q as the center and any suitable radius (keep this radius)

Now, using the same radius, construct arcs 5 & 6 using point P as the center so that they intersect arcs 3 & 4 to get points X & Y

Given a point on a circle, construct the tangent to the circle at the given point .

PROCEDURE:

Given: Point A on circle O.

5

3

X

2

1

O

A

Q

P

Construct arcs 1 and 2 using any suitable radius and A as the center

Y

6

4

1) Draw OA.

2) Find the midpoint M of OA (perpendicular bisector of OA)

3) Construct a 2nd circle with center M and radius MA

5) Draw tangents AX & AY

Given a point outside a circle, construct a tangent to the circle from the given point.

PROCEDURE:

Given: point A not on circle O

X

3

1

M

O

A

4) So you get points of tangency at X & Y where the arcs intersect the red circle

Construct arcs 1& 2 using a suitable radius greater than ½AO

( keep this radius for the next step)

Construct arcs 3& 4 using the same radius

(greater than ½AO)

You get arcs 5 & 6

Y

4

2

7

B

6

3

1

A

C

8

5

4

2

Given a triangle construct the circumscribed circle.

PROCEDURE:

Given: Triangle ABC

1) Construct the perpendicular bisectors of the sides of the triangle and label the point of intersection F.

Bisect segment BC; Using a radius greater than 1/2BC from point C construct arcs 5 & 6

From point B construct arcs 7 & 8 and draw a line connecting the intersections of the arcs

2) Set your compass pointer to point F and the radius to measure FC.

F

3) Draw the circle with center F , that passes through the vertices A, B, & C

radius

Now construct the perpendicular bisector of segment AB and label point F, where the 3 lines meet.

Bisect segment AC; Using a radius greater than 1/2AC from point C construct arcs 1 & 2

From point A construct arcs 3 & 4 and draw a line connecting the intersections of the arcs

B

A

C

Given a triangle construct the inscribed circle.

PROCEDURE:

Given: Triangle ABC

1) Construct the angle bisectors of angles A, B, & C, to get a point of intersection and call it F

2) Construct a perpendicular to side AC from point F, and label this point G.

F

3) Put your pointer on point F and set your radius to FG.

4) Draw the circle using F as the center and it should be tangent to all the sides of the triangle.

G

X

Y

1) Construct ANYRAY from point A that’s not AB

A

C

D

B

3) Draw segment ZB and copy the angle AZB ( 1) to vertices X & Y

Remember you made 3 because you are dividing by 3, but if you wanted to divide by, say, 6 you would have to make 6 congruent parts on the ray and so on for 7,8,9…

Given a segment, divide the segment into a given number of congruent parts.

Given: Segment AB

PROCEDURE:

Divide AB into 3 congruent parts.

2) Construct 3 congruent segments on the ray using ANY RADIUS starting from point A. Label the new points X, Y, & Z

X

So AC=CD=DB

Three congruent segments

Y

1

Use any suitable radius that will give some distance between the points

4) Draw the the rays from X & Y, they should be parallel to the segment ZB and divide AB into 3 congruent parts.

Z

Remember keep the same radius!!

a

b

a

b

c

c

x

x

Given three segments construct a fourth segment (x) so that the four segments are in proportion.

Given:

Construct: segment x such that

b

PROCEDURE:

a

1) Using your straight edge construct an acute angle of any measure.

1

2) On the lower ray construct “a” and then “c “ from the end of “a”.

c

3) On the upper ray construct “b” and thenconnect the ends of “a & b”

4) Next copy angle 1 at the end of “c”and then construct the parallel line

Make sure to set your radius to more

than 1/2NM then:

Mark off 2 arcs from M. Keep the same

radius and mark off 2 more arcs from N,

crossing the first two.

Draw the perpendicular bisector PQ

through point O

a

b

a

x

x

b

2) Bisect a+b (XY) and label point M.

Given 2 segments construct their geometric mean.

Mark off 2 arcs from Y.

Keep the same radius and mark off 2 more arcs from X, crossing the first two.

Given:

Construct: segment x such that:

Mark off 2 equal distances on either

side of point O using any radius and

then bisect this new segment

(I used the distance from O to M,

but remember any radius will do)

The orange segment is x the geometric mean between the lengths of a and b

PROCEDURE:

9

3

1

7

X

Q

K

1) Draw a ray and mark off a+b.

6

5

X

Y

M

3) Construct the circle with center M and radius = MY (or MX)

a

b

O

N

4) Construct a perpendicular where segment “a” meets segment “b” (point O)

8

10

4

2

P

L

If a figure is a locus then it is the set of all points that satisfy one or more conditions.

The term “locus” is just a technical term meaning “a set of points”.

So , a circle is a locus.

Why??

Because it is a set of points a given distance from a given point.

Remember it is a SET OF POINTS so if you recall the idea of sets from algebra it is possible for a set to be empty.

So a set could be:

The empty set. (no points fit the condition or conditions)

A single point.

Two points, three points….

An infinite set of points. (like a line, circle, curve,…)