Lecture 4: Geometry II

Transcription

1 Lecture 4: Geometry II LPSS MATHCOUNTS 19 May 2004 Some Well-Known Pytagorean Triples A Pytagorean triple is a set of tree relatively prime 1 natural numers a,, and c satisfying a = c 2 : = = = = = = = No factors common to all tree numers oter tan 1

2 Area of a Rigt Triangle Recall tat te area of a rigt triangle is easily found y considering te triangle to e alf of a rectangle: Area = 1 2 Area of Aritrary Triangles (Case 1) In tis case, te vertex located aove te ase is in etween te oter two vertices: 1 2 Area = Area 1 + Area 2 = = 1 2 ( ) = 1 2

3 Area of Aritrary Triangles (Case 2) In tis case, te vertex located aove te ase is outside te projection of te ase: 1 2 Area = Area 2 Area 1 = = 1 2 ( 2 1 ) = 1 2 Area of Triangles Summary Te area of any triangle is given y te formula Area = 1 2, were (te ase) is te lengt of any side of te triangle, and (te eigt or altitude) is te perpendicular distance measured from te line containing te ase to te remaining vertex.

4 Special Triangles: Te Triangle Recall tat an equilateral triangle as interior angles of measure 180 /3 = 60. We can otain a rigt triangle y isecting an equilateral triangle: 30 2s s From Pytagoras: 2 + s 2 = (2s) 2 = 4s 2 = 4s 2 s 2 = 3s 2 = 3s 1.732s Area = s = 2 s2 60 Special Triangles: Te or Rigt Isosceles Triangle Recall tat a square as interior angles of measure 90. We can otain a rigt isosceles triangle y isecting a square along its diagonal: 45 s D s 45 From Pytagoras: D 2 = s 2 + s 2 = 2s 2 D = 2s

5 Te Area of a Circle Recall tat te circumference of a circle is C = 2πr, were r is te radius. Consider approximating a circle y an n-sided regular polygon: Area of polygon = = 1 ( ) }{{} 2 }{{} circumference of polygon n terms Area of Circle = 1 n 2 rc = 1 2 r(2πr) = πr 2 Te Area of a Circular Sector We can find te area y noting tat te area of te sector is proportional to te lengt of sutended arc: r s Sector Area = s (Circle Area) 2πr = s 2πr πr 2 = 1 2 sr Tis is easy to rememer ecause it is te same as a triangle of ase s and eigt r.

6 Te Area of a Trapezoid A trapezoid is a quadrilateral wit exactly one pair of opposite sides parallel. We find te area y decomposing te trapezoid into rigt triangles and a rectangle: x 1 y y 2 z Trapezoid area = 1 2 x + y z = ( 1 2 x y ) + ( 1 2 y + 1 ) 2 z = 1 2 (x + y) (y + z) = = Te Area of a Parallelogram A parallelogram is a quadrilateral wit exactly two pairs of opposite sides parallel (and congruent). We find te area using te same metod as for te trapezoid. Note tat any side of te parallelogram can e used as te ase. a 1 2 a Parallelogram area = 1 = a 2

7 Pyramids A pyramid is a solid figure wic as a polygon for its single ase and triangles for its sides (or faces). Examples: Triangular Pyramid Rectangular Pyramid Pyramid Surface Area and Volume Te surface area of a pyramid is found y summing te areas of all of te triangular faces, plus te area of te ase. Te volume of a pyramid is given y te formula V = 1 3 A were A is te area of te ase, and is te perpendicular eigt of te free vertex aove te ase. (Note: same formula works for a cone!)

8 Pyramid Volume Example A rectangular pyramid as a ase measuring 6 inces y 10 inces. If te eigt is 40 inces, wat is te volume of te pyramid in cuic inces? Solution: V = 1 3 A = (6 10) = = = 800 cuic inces. Prisms A prism is a solid figure consisting of two identical, parallel ases, connected y faces consisting of rectangles: Triangular Prism Rectangular Prism

9 Prism Surface Area and Volume Te surface area of a prism is found y summing te areas of all of te rectangular faces, plus te area of te two identical ases. Te volume of a prism is given y te formula V = A were A is te area of te ase, and is te perpendicular distance etween te ases. (Note: same formula works for a circular cylinder!) Prism Volume Example A triangular prism as a rigt triangle ase measuring 3 inces y 4 inces y 5 inces. If te eigt is 40 inces, wat is te volume of te prism in cuic inces? Solution: V = A = = 6 = 6 40 = 240 cuic inces. 2

10 Cords of Circles A cord is a line segment joining two points on a circle: Cords of Circles (Cont.) Note tat te center of te circle lies on te perpendicular isector of te cord. Tus, given two cords, we can find te center:

11 Rigt Triangle Inscried in a Circle We can find te center, and tus te radius, y locating te perpendicular isectors of te two legs. a Rigt Triangle Inscried in a Circle (Cont.) We can find te center, and tus te radius, y locating te perpendicular isectors of te two legs. Note tat we ave constructed a rectangle of sides a/2 and /2 wose diagonal is te circle s radius of lengt r = D/2, were D is te diameter of te circle.

12 Rigt Triangle Inscried in a Circle (Cont.) From Pytagoras, ten (a/2) 2 + (/2) 2 = (D/2) 2 = a = D 2 so tat te ypotenuse of te triangle must e a diameter of te circle! Summary: Any rigt triangle inscried in a circle forms a diameter of te circle wit its ypotenuse! Some formulas for Area and Volume 1. Area of a triangle given te ase and eigt : A = /2 2. Area of square, given te side lengt s: A = s 2 3. Area of square, given te diagonal lengt d: A = d 2 /2 4. Area of a romus, given te lengt of te diagonals d 1 and d 2 : A = d 1 d 2 /2 5. Area of a circle given te radius r: A = πr 2 6. Area of a trapezoid given ases 1, 2, and te eigt : A = ( )/2 7. Volume of cylinder or prism given ase area B and eigt : V = B 8. Volume of cone or pyramid given ase area B and eigt : V = B/3 9. Volume of spere given radius r: V = 4πr 3 /3 10. Surface area of spere given radius r: A = 4πr 2

13 Te numer of diagonals in a regular polygon A diagonal of a polygon is a line segment joining any two nonadjacent vertices. Let us count te numer of diagonals in a few different regular polygons. We will use n to denote te numer of sides in te polygon and N n to denote te numer of diagonals: Te square (n = 4): N 4 = = 2 Te pentagon (n = 5): N 5 = = 5 Te exagon (n = 6): N 6 = = 9 Te numer of diagonals in a regular polygon (Cont.) Te eptagon (n = 7): N 7 = = 14 Te general n-sided polygon: N n = (n 3) + (n 3) + (n 4) + (n 5) (n 3)(n 2) = (n 3) + T n 3 = n [ ] [ ] (n 2) 2 (n 2) = (n 3) 1 + = (n 3) = (n 3) 2 + n 2 2 n(n 3) = 2

14 Example Prolem Solved Using Numer of Diagonals Prolem Suppose you are a gym teacer wit 35 students in your ping-pong class. You need to set up a round-roin tournament in wic every student plays a matc against every oter student. How many matces sould e sceduled? Solution Consider placing eac student at te vertex of a 35-sided regular polygon. Asking for te numer of matces is equivalent to asking te numer of diagonals plus te numer of sides of te polygon. We ave 35(35 3) N 35 = = = To tis numer we must add te numer of sides on te polygon: # Matces = = 595