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# College Geometry MATH 323

ISU

GPA 3.92

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This 10 page Study Guide was uploaded by Emerald Quitzon on Monday October 12, 2015. The Study Guide belongs to MATH 323 at Indiana State University taught by Elizabeth Brown in Fall. Since its upload, it has received 33 views. For similar materials see /class/222209/math-323-indiana-state-university in Mathematics (M) at Indiana State University.

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Date Created: 10/12/15

Construction Reference Construct the perpendicular bisector of a line segment 0r construct the midpoint of a line segment 1 Begin with line segment XY X 2 Place the compass at pointX Adjust the compass radius so that it is more than 12XY Draw two arcs as shown here X 3 Without changing the compass radius place the XA compass on point Y Draw two arcs intersecting the previously drawn arcs Label the intersection points A and B X X3 4 Using the straightedge draw line AB Label the intersection point M PointM is the midpoint of line A segment X Y and line AB is perpendicular to line segment X Y X M B Given point P on line k construct a line through P perpendicular to k 1 Begin with line k containing point P N Place the compass on point P Using an arbitrary radius draw arcs intersecting line k at two points Label the intersection points X and Y E Place the compass at pointX Adjust the compass radius so that it is more than 12XY Draw an arc as shown here 4 Without changing the compass radius place the compass on point Y Draw an arc intersecting the previously drawn arc Label the intersection pointA 5 Use the straightedge to draw line AP Line AP is perpendicular to line k Given point R n0t on line k construct a line through R perpendicular to k 1 Begin with point line k and point R not on the line R k gt 2 Place the compass on point R Using an arbitrary radius R k draw arcs intersecting line k at two points Label the X Y intersection points X and Y 3 Place the compass at pointX Adjust the compass radius 0R k so that it is more than 12XY Draw an arc as shown X Y here 4 Without changing the compass radius place the 0R k compass on point Y Draw an arc intersecting the X Y previously drawn arc Label the intersection point B gtltB 5 Use the straightedge to draw line RB Line RB is perpendicular to line k Construct the bisector of an angle 1 Let pointP be the vertex of the angle Place the compass on point P and draw an arc across both sides of the angle Label the intersection points Q and R 2 Place the compass on point Q and draw an arc across the interior of the angle 3 Without changing the radius of the compass place it on point R and draw an arc intersecting the one drawn in the previous step Label the intersection point W 4 Using the straightedge draw ray PW This is the bisector of AQPR Construct an angle congruent to a given angle 1 To draw an angle congruent to 4A begin by drawing a ray with endpoint D 2 Place the compass on pointA and draw an arc across both sides of the C angle Without changing the compass radius place the B compass on point D and draw a long arc crossing the ray Label the three intersection points as shown 3 Set the compass so that its radius is BC Place the compass on point E C and draw an arc intersecting the one F drawn in the previous 3 step Label the intersection point F A D E 4 Use the straightedge to draw ray DF c LEDF E ABAC F B A D E Given a line and a point construct a line through the point parallel to the given line 1 Begin with pointP and line k P k 2 Draw an arbitrary line through point P intersecting line k Call the intersection point Q Now the task is to construct an angle with vertex P P congruent to the angle of intersection Q k 3 Center the compass at point Q and draw an arc intersecting both lines Without changing the radius of the compass center it at pointP and P draw another arc Q k 4 Set the compass radius to the distance between the two intersection points of the rst arc Now center the compass at the point where the second arc intersects line PQ Mark the arc intersection point R V39 Line PR is parallel to line k Given a line segment as one side construct an equilateral triangle This method may also be used to construct a 60 quotangle 1 Begin with line segment TU U 2 Center the compass at point T and set the compass radius to TU Draw an arc as shown U 3 Keeping the same radius center the compass at V point U and draw another arc intersecting the rst one Let point Vbe the point of intersection 4 Draw line segments TVand UV Triangle TU V is an equilateral triangle and each of its interior angles has a measure of 60 Divide a line segment into n congruent line segments In this example n 5 Begin with line segment AB It will be divided into ve congruent line segments Ao og N Draw a ray from pointA Use the compass to step off ve uniformly spaced points along the ray Label the last point C E Draw an arc with the compass centered at point A with radius BC Draw a second arc with the compass centered at point B with radius AC Label the intersection point D Note that ACBD is a parallelogram 4 Use the compass to step off points along line segment DB using the same radius that was used for the points along line segment AC V39 Use the straightedge to connect the corresponding points These line segments will be parallel They cut line segments AC and DB into congruent segments Therefore they must also cut line segmentAB into congruent segments Given a circle its center point and a point on the exterior of the circle construct a line through the exterior point tangent to the circle Begin with a circle centered on point C Point P is on the exterior of the circle P N E Center the compass on point M Draw a circle through points C and P It will intersect the other circle at two points R and S 4 Points R and S are the tangent points Lines PR and PS are tangent to the circle centered on point C Draw line segment CP and construct point M the midpoint of line segment CP For the construction of the midpoint refer to the perpendicular bisector construction on page 1 P M R P s Construct the center point of a given circle 1 Begin with a circle but no center point 2 Draw chord AB E Construct the perpendicular bisector of chord AB Let C and D be the points where it intersects the circle Refer to the construction of a perpendicular bisector on page 1 4 Chord CD is a diameter of the circle Construct point P the midpoint of diameter CD Point P is the center point of the circle Refer to the construction of the midpoint of a line segment on page 1 Given three n0nc0llinear points construct the circle that includes all three points 1 Begin with points A B and C A C D C C 2 Draw line segmentsAB and BC A C B 3 Construct the perpendicular bisectors of line 0 segments AB and BC Refer to the perpendicular bisector construction on page 1 Let point P be the intersection of the perpendicular bisectors B 4 Center the compass on point P and draw the circle through points A B and C Given a triangle circnmscribe a circle 1 Begin with triangle STU N If a circle is circumscribed around the triangle then all three vertices will be points on the circle so follow the instructions above for construction of a circle through three given points Given a triangle inscribe a circle 1 Begin with triangle KLM N Construct the bisectors of K and AL Refer to the angle bisector construction on page 4 Let point Q be the intersection of the two angle bisectors E Construct a line through point Q perpendicular to line segment KL Let point R be the point of intersection Refer to the construction of a perpendicular line through a given point on page 3 4 Center the compass on point Q and draw a circle through point R The circle will be tangent to all three sides of a triangle

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