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15.7 Constructing a parallelogram Construct a parallelogram given the lengths of two adjacent sides a and b and of a diagonal d (Fig. 15-21).
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Fig. 15-21
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Solution
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Three vertices of the parallelogram are obtained by constructing ^ABD by construction 7. The fourth vertex, C, is obtained by constructing ^BCD upon diagonal BD by construction 7. Vertex C may also be obtained by constructing BC AD and DC AB.
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CHAPTER 15 Constructions
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15.6 Circle constructions
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CONSTRUCTION
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14: To construct a tangent to a given circle through a given point on the circle
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Given: Circle O and point P on the circle (Fig. 15-22) To construct: A tangent to circle O at P 4 4 4 Construction: Draw radius OP and extend it outside the circle. Construct AB ' OP at P. AB is the required tangent. (A line perpendicular to a radius at its outer extremity is a tangent to the circle.)
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Fig. 15-22
Fig. 15-23
CONSTRUCTION
15: To construct a tangent to a given circle through a given point outside the circle
Given: Circle O and point P outside the circle (Fig. 15-23) To construct: A tangent to circle O from P Construction: Draw OP, and make OP the diameter of a new circle Q. Connect P to A and B, the intersections of circles O and Q. Then PA and PB are tangents. (jOAP and jOBP are right angles, since angles inscribed in semicircles are right angles.)
CONSTRUCTION
16: To circumscribe a circle about a triangle
Given: ^ABC (Fig. 15-24) To construct: The circumscribed circle of ^ABC Construction: Construct the perpendicular bisectors of two sides of the triangle. Their intersection is the center of the required circle, and the distance to any vertex is the radius. (Any point on the perpendicular bisector of a segment is equidistant from the ends of the segment.)
Fig. 15-25 Fig. 15-24
CONSTRUCTION
17: To locate the center of a given circle
Given: A circle (Fig. 15-25) To construct: The center of the given circle Construction: Select any three points A, B, and C on the circle. Construct the perpendicular bisectors of line segments AB and AC. The intersection of these perpendicular bisectors is the center of the circle.
CHAPTER 15 Constructions
18: To inscribe a circle in a given triangle
CONSTRUCTION
Given: ^ABC (Fig. 15-26) To construct: The circle inscribed in ^ABC Construction: Construct the bisectors of two of the angles of ^ABC. Their intersection is the center of the required circle, and the distance (perpendicular) to any side is the radius. (Any point on the bisector of an angle is equidistant from the sides of the angle.)
Fig. 15-26
SOLVED PROBLEMS
15.8 Constructing tangents A secant from a point P outside circle O in Fig. 15-27 meets the circle in B and A. Construct a triangle circumscribed about the circle so that two of its sides meet in P and the third side is tangent to the circle at A.
Solution
Use constructions 14 and 15: At A construct a tangent to circle O. From P construct tangents to circle O intersecting the first tangent in C and D. The required triangle is ^PCD.
Fig. 15-27
15.9 Constructing circles Construct the circumscribed and inscribed circles of isosceles triangle DEF in Fig. 15-28.
Solution
Use constructions 16 and 18. In doing so, note that the bisector of jE is also the perpendicular bisector of DF. Then the center of each circle is on EG. I, the center of the inscribed circle, is found by constructing the bisector of jD or jF. C, the center of the circumscribed circle, is found by constructing the perpendicular bisector of DE or EF.
CHAPTER 15 Constructions
Fig. 15-28
15.7 Inscribing and Circumscribing Regular Polygons
CONSTRUCTION
19: To inscribe a square in a given circle
Given: Circle O (Fig. 15-29) To construct: A square inscribed in circle O Construction: Draw a diameter, and construct another diameter perpendicular to it. Join the end points of the diameters to form the required square.
Fig. 15-29
CONSTRUCTION
20: To inscribe a regular octagon in a given circle
Given: Circle O (Fig. 15-30) To construct: A regular octagon inscribed in circle O Construction: As in construction 19, construct perpendicular diameters. Then bisect the angles formed by these diameters, dividing the circle into eight congruent arcs. The chords of these arcs are the sides of the required regular octagon.
Fig. 15-30
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