Math 310  Fall 20 06  Test #2  100 points  ANSW ERS ! 1a. The number of distinct points necessary to determine a specific line is 2“Two distinct points determine a line.” Ie “through any two points there is one and only one line.”How many lines can YOU draw t hrough the tw o points at right ? CHow many lines can be drawn t hrough one single point? C b. The number of planes containing a line is infinite.Consider the line at the corner of the room, where two walls meet. Each wall represents a different plane through the line. A w all could be built out f rom that corner in any direction. c. The number of intersection points of a pair of skew lines is 0 (none).Skew lines never meet; they exist in different planes. An example of skew lines: a line where the west wall of the room meets t he ceiling, and the line wherethe north w all meets the floor. d. The number of points shared by a plane and a line perpendicular to the plane is one(where the line pierces the plane, as a needle might pierce a piece of paper) e. The number of non-collinear points needed to determine a plane is 3Three non-collinear points det ermine a plane. That ’s why a three-legged stool can sit solidly on a floor, w hile a two-legged stool w ould fall over... andw hy a four-legged chair sometimes rocks (when the fourth leg does not end in the plane determined bythe first three leg bottoms). f. The number of planes containing tw o lines which intersect is one.The point w here the lines intersect, together with one additional point from each line, comprises t hreenoncollinear points, enough to determine a plane. Then, because tw o points of each line belong to theplane, each ent ire line must lie in the plane. g. The number of different (non-congruent) triangles with sides 5cm, 7cm, & 9cm, is 1.There is only one size & shape triangle having sides 5cm, 7cm, & 9cm. By t he “ SSS” theorem of trianglecongruence, there can be only one (all others are congruent). A nd since 9 # 5+ 7, these values satisfythe t riangle in equality, and so there is such a triangle. h. The number of different (non-congruent) triangles with sides 2cm, 4cm, & 6cm, is 0 (n one) .You might say to yourself, t his w as a trick question. But in fact, having this question right after part gabove w as int ended t o make you st op and think (w hy are there tw o of these questions?). What isdiff erent here is the f ailure to meet the requirements of t he triangle inequality. To hav e a genuine triangle,surrounding some interior points, t he t w o short est segments must tot al more t han the longest segment.2cm 4cm When w e attempt to build a triangle using these6cm lengths, w e see it never gets off the ground! The 2cm & 4cm segments are just enough to span 6cm when laid flat. We need more length to “ tent up” these segments and have them meet!Math 310  Fall 20 06  Test # 2  10 0 points  ANSW ERS ! Continued. .. 2. Of t he choices given the BEST completion of each statement f ollow s:A circle B cube C hexahedron D line E octahedron F parallelogramG plane H point I polygon J polyhedron K prism L pyramidM rectangle N rhombus O segment P sphere Q square R simple closedcurvea. A simple closed curv e consist ing of line segments is a POLYGON.b. The polyhedron illust rat ed at right is an OCTAHEDRON. 2bc. The set of all points in a plane equally distant from a given point P is a CIRCLE.d. The f igure at right can be folded up int o the polyhedron know n as a PYRAM ID. 2de. A quadrilat eral w it h all sides congruent is a RHOMBUS.f. A parallelogram w ith an interior angle measuring 90° is a RECTA NGL E. 3. In the illust ration at right , P & Q are tangled w ith a simple closed plane curve (SCPC).Are t he point s P and Q on the same side of t he c urve? We can t ell t hat P & Q are on opposite sides of the curve in several w ays.Probably the most direct is t o connect them w it h a segment (OR a curve) and see that t he segment crosses the given SCPC an odd number of t imes,show ing, by the Jordan Curve Theorem, t hat P & Q are on opposit e sides–one is int erior and t he ot her exterior t o t he curve.CAnother, similar method: Connect P to an exterior point. The connecting curve crosses the given SCPC an odd number of times, showing that P is inthe INTERIOR of the curve. Similarly, Q is connected to an exterior point, and t he even number of SCPC crossings show s us Q is EXTERIOR.Another method is to color in the interior of t he curve (pick an interior Cpoint to start, and obey the lines)... Or color the exterior in. Try it on the copy at right. 4. Find the measure of the angle marked x. x 72° 1 Angle 2 is t he supplement of the 135 ° angle,which makes m(p2) = 180° – 135° = 45° .m(px) = sum of the t w o remot e int erior angles, 7 2°+ 45 °= 117° 2 ²135° But most people would probably find m(p1):m(p1) = 180° – (72°+ 45°) = 180° – 117° = 63°then find m(px)= 180° – m((p1) = 180° – 63° = 117° 5. Given the triangle and measurements illustrated, find the measure of t he angle marked y.Explain w hat triangles are similar and how you know they are.A ªADE ~ ªABC because the sides of ADE are cut into segmentsof the same proportions (6:9 as 4:6 — both are in the ratio 2:3) 6 60° 4 w hich tells us AC is parallel to DE, t hus f orming the same angles.(This can also be argued on the basis of SAS: A C:AD is 6:15 and x 78°B C AC:A E is 4:10 – both reduce to the rat io 2:5 betw een those sides,angle A is shared, therefore congruent.) 9 6Knowing those tw o triangles are similar tells us angle D (marked y) y Is congruent to angle x, w hich must be 180° – ( 60° + 78°) = D E180° – 138° = 42°Math 310  Fall 20 06  Test # 2  10 0 points  ANSW ERS ! Continued. .. 6. Without using a protractor, showing your w ork, f ind the sum of the measures of t he interior angles in the polygon at right: " $m(p") + m(p$) + m(p() + m(p*) + m(p,) + m(p2) = 4* 180°= 720° ( 2 We triangulate the region inside the polygon....and see four triangles covering the region, while …