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Geometry SOL “Things to Know:”

y. x. rise. -1 , 1. run. -10. -3. 0. 5. x 2 +x 1 y 2 +y 1 Midpt = -------, ------- 2 2. 3 + -5 4 + -2 -------, ------- 2 2. -2 2 -----, ----- 2 2. A. D. C. B.

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Geometry SOL “Things to Know:”

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  1. y x rise -1 , 1 run -10 -3 0 5 x2+x1 y2+y1 Midpt = -------, ------- 2 2 3 + -5 4 + -2 -------, ------- 2 2 -2 2 -----, ----- 2 2 A D C B Geometry SOL“Things to Know:” • G.1 Hypothesis and Conclusions (if … then …) Examples p  q (original) If today if Friday, then tomorrow is Saturday. q  p (converse) If tomorrow is Saturday, then today is Friday. ~p  ~q (inverse) If today is not Friday, then tomorrow is not Saturday. ~q  ~p (contrapostive) If tomorrow is not Saturday, then today is not Friday. ~ means “not” Law of Syllogism: If p  q is true and q  r is true, then p  r is true. Example:(pq) If fossil fuels are burned, then acid rain is produced. (qr) If acid rain is produced, then wildlife suffers. (pr) if fossil fuels are burned, then wildlife suffers. • G.2 Distance (d) = √(x2-x1)2 + (y2-y1)2 Midpoint (midpt): ½ sum (an average) Examples: Find the distance between (2,5) and 3,10) Find the midpt of AB d = √(3-2)2 + (10-5)2 d = √12 + 52 d = √1 + 25 = √26 = 5.099midpt = ½ (-10+5) = ½ (-5) = -2.5 Ex: Find the length of AB Ex: Find midpt of AB d = √(x2-x1)2 + (y2-y1)2 = √(3-(-5))2 + (4-(-2))2 = √(3+5)2 + (4+2)2 = √(8)2 + (6)2 = √64 + 36 = √100 = 10 B (3,4) D (7,1) A (-5,-2) C (4,-3) rise y2 - y1 Slope = ------- or m = -------- run x2 - x1 y2 - y1 1 - (-3) Ex: Find the slope of CDm = -------- = ---------- x2 - x1 7 – 4 4 m = ----- 3 y2 - y1 7 – 1 6 Ex: Find the slope of the line going through (1,1) and (6,7) Slope: m = -------- = ------ = --- x2 - x1 6 – 1 5

  2. t 1 2 m 3 4 5 6 n 7 8 t C A 3x + 8 6 5x – 2 B m 5x + 12 4 5 D 2 3x + 8 E n 3 1 Vertical angles are congruent () • G3 Complimentary angles: 2’s whose sum = 90° Supplementary angles: 2’s whose sum = 180° Corresponding ’s, alternate interior ’s, and alternate exterior ’s are  when parallel lines are cut by a transversal Consecutive ’s are supplementary when parallel lines are cut by a transversal lines m // n, line t is a transversal Corresponding ’s: 1&5, 3&7, 2&6, 4&8 Alternate Int ’s: 3&6, 4&5 Alternate Ext ’s: 1&8, 2&7 Consecutive ’s: 3&5, 4&6 Vertical ’s: 1&4, 2&3, 5&8, 6&7 Sum of interior angles in a polygon: S = (# of sides – 2)*180° Sum of exterior angles in a polygon: Always 360° Each exterior angle in the polygon: 360 ÷ n (# of sides or # of angles) Each interior angle in the polygon: 180 – exterior  (linear pairs are supplementary) Examples: Find mABD, mCBE, mABC Find m1, m2, m3, m4, m5, m6 5x+12 + 3x+8 = 180° (linear pair) 5x – 2 = 3x + 8 (corresponding ’s ) 8x + 20 = 180 -3x -3x -20 -20 2x – 2 = 8 8x = 160 + 2 + 2 x = 20° 2x = 10 x = 5 mABD = 5(20) + 12 = 112° mCBE = 112° (vertical  with ABD) 5(5) – 2 = 23° mABC = 3(20) + 8 = 68° m1 = 157 (vertical  to 2) m2 = 157° (alt int  with 5) m3 = 23° (corresponding to 4) m4 = 23° (vertical  with 5x-2) m5 = 157° (corresponding to 1) m6 = 157° (linear pair)

  3. C K A B L J J A 15 9 5 3 40° 40° C L B K 4 12 t 150° Is a // b? Yes Why? Corresponding ’s are  when two parallel lines are cut by a transversal! • G3 examples cont For a regular (all sides  and angles ) octagon find: • Sum of interior angles: S = (n-2)*180 = (8-2)*180 = 6*180 = 1080° • Sum of exterior angles: 360° • each exterior angle: 360 ÷ 8 = 45° • each interior angle: 180 – 45 = 135° • G4 example • G5 Ways to prove ∆’s  (congruent – copies, all corresponding ’s & sides ) • SSS • SAS • ASA • AAS • (additional ways for right ∆’s only) • HL – hypotenuse-leg (also SSS after use of Pythagorean Thrm) • HA – hypotenuse-angle (also AAS) • LA – leg-angle (also AAS or ASA depending on which leg ) • LL – leg-leg (also SAS with 90° angle) Ways to prove ∆’s ~ (similar – angles , all sides proportional) • AA (third angle must be  because all add to 180°) • SSS • SAS a 150° b If AC  JL (side), A  J (included angle) and AB  JK (side) then ∆ABC  ∆JKL (SAS) ∆ABC ~ ∆JKL AB * scaling factor = JK 3 * 3 = 9 AC * scaling factor = JL 5 * 3 = 15 C 12 8 A 16 B SSS C Y 3 12 2 X 4 Z A 16 B SAS Y 3 X 4 Z

  4. 13 x x 2 12 5 B 120° 40° C 20° A 45° y 5 45° x y 15 y 60° 60° 60° 10 x x 5 y 30° 30° 30° x 7 y x • G6 examples The sum of two sides of a triangle must be greater than the third side. Is this triangle possible? 1, 2, 3 Why not? Because 1+2 is not greater than 3 Are the following triangles possible? 5, 6, 7 Yes 5+6 > 7 3, 4, 5 Yes 3+4 > 5 (also a Pythagorean triple) 7, 10, 11 Yes 7+10 > 11 2, 6, 3 No 2+3 < 6 The longest side is opposite the largest ; smallest side is opposite the smallest  What is the shortest side? BC What is the longest side? AB What if the middle side? AC • G7 Pythagorean Theorem (c2 = a2 + b2) Solve for x 52 + 22 = x2 x2 + 122 = 132 25 + 4 = x2 x2 + 144 = 169 29 = x2 - 144 = -144 √29 = x x2 = 25 x = 5 Is 3, 4, 5 a right triangle? Is 11, 17, 23 a right triangle? Yes, because 32 + 42 = 52 No, because 112 + 172≠ 232 9 + 16 = 25 121 + 289 ≠ 529 25 = 25 650 ≠ 529 Special Case Right Triangles 30-60-90 45-45-90 Side opposite 30°: ½ hypotenuse Side opposite 45°: ½ hypotenuse √2 Side opposite 60°: ½ hypotenuse √3 hyp = leg * √2 short leg = long leg ÷ √3 leg = hyp ÷ √2 long leg = short leg * √3 short leg = hyp ÷ 2 hyp = short leg * 2 x = ½*15 = 7.5 y = 7.5√3 x = 5 y = 5√2 x = 7÷√3 7 * √3 7√3 x = ----- ----- = -------- √3 * √3 3 7√3 2 14√3 y = ----- * --- = -------- 3 1 3 x = 10÷√2 10 * √2 10√2 x = ----- ----- = -------- √2 * √2 2 x = 5√2 y = 5√2 y = 2*5 = 10 x = 5√3

  5. trig function sides of ∆ opposite tan x = ------------------ adjacent angle Inversetrig function sides of ∆ opposite tan-1 --------------- = x adjacent Always press 2nd then the function when finding angles angle 30° x B C 50x+5 5 20x+65 D A 5 3 x° 4 opposite sin x = ------------------- hypotenuse adjacent cos x = ------------------ hypotenuse opposite tan x = ------------------ adjacent • G7 continued Trigonometry Some old hippy caught another hippytripping on acid. or SOH CAH TOA • G8 Quadrilaterals Properties of a parallelogram • Opposite sides  • Opposite ’s  • Consecutive ’s are supplementary (=180°) • Diagonals bisect each other • Opposite sides parallel Properties of a Rectangle 1-5 same as a parallelogram • 4 right ’s • Diagonals  Properties of a Rhombus 1-5 same as a parallelogram • 4  sides • Diagonals bisect opposite ’s • Diagonals perpendicular Properties of a Square All of the above properties of parallelogram, rectangle, and rhombus • G.9 Tessellations The sum of all angles around a point is 360° A regular polygon will tessellate a plane if it’s interior angle will ÷ into 360° evenly. 5 tan 30° = ----- x x tan 30° = 5 5 x = ---------- = 8.66 tan 30° 3 4 3 sin x = ---- cos x = ---- tan x = ----- 5 5 4 sin-1 cos-1 tan-1 2nd sin (3/5) 2nd cos (4/5) 2nd tan (3/4) x≈ 37° x ≈ 37° x ≈ 37° 9 9 9 9 Will a regular hexagon tessellate a plane? ext  = 360°÷6 = 60° int  = 180° - 60° = 120° Will 120 go into 360 with no remainder? yes Will a regular octagon tessellate a plane? ext  = 360°÷8 = 45° int  = 180° - 45° = 135° Will 135 go into 360 with no remainder? no!!

  6. 200° x 60° x x 100° 30° x 100° 120° x x 5 3 10 3 2 10 x 5 60° 50° x x 240° 20° x x = 12 x 120° 60° 5 12 Central x = 60° Inscribed x = 100° Exterior  x = 50°-20° = 30° = 15° 2 2 Exterior  x = 240°-120° = 120° = 60° 2 2 • G.10 Circles Chords, Secants and Tangents Area of a sector (% of total area) θ A = -------- * πr2 where θ is central angle 360° Interior x = 100°+120° = 220° = 110° 2 2 Inscribed x = ½(60°)x = (30°) Exterior x = 100°-30° = 70° = 35° 2 2 2 “outside part times whole thing = tangent squared” 3(3 + x) = 102 9 + 3x = 100 3x = 91 x = 30.33 to get “x” 5*x = 2*10 5x = 20 x = 4 “outside part times whole thing ” 3(3 + x) = 2*(2 + 5) 9 + 3x = 14 3x = 5 x = 5/3 θ A = ------ * πr2 360° 60° 25π A = ------ * π52 = ------ ≈ 13.083 360° 6 “tangents = from same point” x2 = 122 x = 12

  7. arc length 60° 5 θ 60° 1 10π A = ------ * 2πr = ------- * 2π5 = ----* 10 π = ------ ≈ 5.236 360° 360° 6 6 G.10 Circles continued Length of an Arc (% of Circumference) θ Arc Length = -------- * 2πrwhere θ is central angle 360° • G.11 Constructions (see pages in your book for figures) • Angle Bisector (pg 32 - 33) • Place compass point at the vertex B. With the center at B, draw an arc that intersects the sides of the angle. Label the point of intersection X and Y. • Place the compass point at X and draw an arc in the interior of ABC. Place the compass point at Y. Using the same radius, draw an arc that intersects the previous arc. • Label the point of intersection Z. Draw BZ, which is the angle bisector of ABC. ABZ  CBZ. • Congruent Angle to a Given Angle (pg 31) • Top copy ABC, place the compass point at the vertex, B. With the center at B, draw an arc that intersects the sides of the angle. Label the points of intersection D and E. • Draw a new line m and mark a point P on the line. Place your compass point at P. Use the same radius that you used in step 1, and draw an arc with its center at P that intersects line m. Label the intersection point Q. • On the original angle, measure arc ED with your compass (see figure 3). Then, on line m, place your compass point at Q. Use a radius equal to arc ED, draw an arc with center at Q. Lavel the intersection point of the two arcs R. • Draw PR. The measure of PQR is the same as the measure of ABC. • Perpendicular Bisector (pg 24, 44, 236) • Place compass point on point A. Use a radius that is more than half the length of AB. Draw an arc that intersects AB. • Using the same radius, place the compass point at B, and draw an arc that intersects the previous arc BOTH above and below AB. Label the intersection points of the arcs P and Q. • Draw line PQ, which is the perpendicular bisector of AB. The diagram shows PW as a line which intersects AB. You can use this construction to find the midpoint of any line segment.

  8. 2 x 7 5 • Perpendicular Thru a Point on a Line (pg 44) • To draw a perpendicular to line m at point P, first place the compass point at P. Draw 2 arcs of the same radius, one on either side of P, that intersect line m. Label the points of intersection A and B. • Place the compass point at A. Use a radius greater than the previous one, and draw an arc above with center at A. Place the compass point at B. Using the same radius, draw an arc with the center at B that intersects the previous arc. • Label the point of intersection of the two arcs Q. Draw PQ, which is perpendicular to m. • Perpendicular to a Line From a Point Off the Line (pg 44) • To draw a perpendicular to line m through point P, first place the compass at point P. Draw an arc that intersects line m at 2 points. • Label the intersection points S and T. Place the compass point at T. Using any radius longer than half the distance between S and T, draw and arc with center at T on the opposite side of the line from P. • Repeat this same step from point S, using the same radius. Label the intersection point of the 2 arcs Q. Draw PQ, which is perpendicular to line m. • G.12 Making a model of a 3-D figure from a 2-D drawing • G.13 Surface Area & Volume * remember Area is measured in square units Volume is measured in cubic units • G.14 Proportional reasoning Remember: tree to shadow tree to shadow 30 ft x shadow shadow 15 ft 7 ft Area of a RhombusA = ½ diag * diag 30 x --- = ---- 15 7 15x = 210 x = 14 2 x --- = ---- 5 7 5x = 14 x = 14/5 = 2.8

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