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Chapter 10

Chapter 10. Constructions. Given:. A. B. Construction #1. 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.

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Chapter 10

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  1. Chapter 10 Constructions

  2. Given: A B Construction #1 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

  3. Construction #2 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

  4. Construction #3 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.

  5. Construction #4 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.

  6. Construction #5 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.

  7. Construction #6 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.

  8. Construction #7 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.

  9. ConcurrentLines If the lines are Concurrent then they all intersect at the same point. The point of intersection is called the“point of concurrency”

  10. What are the 4 different types of concurrent lines for a triangle? • Perpendicular bisectors • Angle bisectors • Altitudes • Medians • Circumcenter • Incenter • Orthocenter • Centriod Concurrent Lines Point of Concurrency Circumcenter SP Orthocenter SP Incenter SP Centroid SP

  11. 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 Construction #8 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

  12. 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 Construction #9 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

  13. 7 B 6 3 1 A C 8 5 4 2 Construction #10 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

  14. B A C Construction #11 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

  15. 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… Construction #12 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!!

  16. a b a b c c x x Construction #13 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

  17. 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. Construction #14 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

  18. The Meaning of Locus 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.

  19. What can a locus be?? 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,…)

  20. Examples of a Locus in a Plane

  21. Locus Problems

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