Quadrilaterals Here are some examples using quadrilaterals

Transcription

1 Qudrilterls Here re some exmples using qudrilterls

2 Exmple 30: igonls of rhomus rhomus hs sides length nd one digonl length, wht is the length of the other digonl? 4 -

3 Exmple 31: igonls of prllelogrm Given prllelogrm whose sides mesure nd, nd one digonl is c, wht is the length of the other digonl? c + -c simple enough result, ut cn you derive it? I used the cosine rule, ut cn you do it y Pythgors lone?

4 Exmple 3: igonls of Kite ontinuing in this theme, if we hve kite whose non-xis digonl is length c, nd whose sides re length nd, wht is the length of the other digonl? c + - c + +c -c +c -c n you see the similrity to the prllelogrm?

5 Wht out the non-convex kite with the sme side lengths nd digonl? c +c -c - +c -c How out if we re specified the xis, wht is the non-xis digonl? d + d - 4 d -d d

6 This is very similr to the eqution of the ltitude of tringle. Why?

7 Exmple 33: yclic Qudrilterl qudrilterl is inscried in circle of rdius. 3 sides of the qudrilterl hve length. Wht is the length of the fourth side? E Wht is the reltionship etween nd when the fourth side hs length?

8 Exmple 34: Right Trpezoid h +(-+) h

9 Exmple 35: Trpezium: res of Qudrilterls h h (+) Kite:

10 Prllelogrm: θ sin(θ)

11 Exmple 36: res of tringles in trpezoid One pir of tringles formed y the digonls of trpezoid re equl in re. h (+) h (+) h

12 The other pir re not: h (+) h h (+)

13 Exmple 37: imeter of the circumcircle Here is digrm which llows us to find the dimeter of the circumcircle of the tringle whose sides hve lengths, nd c. Why? c c ++c -(+-c) (-+c) (--c)

14 Exmple 38: Qudrilterl with perpendiculr digonls nd one right ngle Qudrilterl hs 3 sides length,, c nd right ngle. It s digonls re perpendiculr. Wht is the length of the remining side? - + +c c

15 Exmple 39: Finding the dimeter of n rc given the perpendiculr offset from the chord. Here is the digrm. Verify tht it is correct: h E h+ 4 h

16 Exmple 40: Npoleon s Theorem Npoleon s Theorem sttes tht if you tke generl tringle nd drw n equilterl tringle on ech side, then the tringle formed y joining the incenters of these new tringles is equilterl. You cn see tht the length is symmetricl in,,c nd hence identicl for the three sides of the tringle. P O Q P E c S c c T c c +-c -+c -++c 6 F

17 Exmple 41: n Isosceles Tringle Theorem is isosceles. E=. We show tht EF=F. c E 4 +c - c 4 F c 4 +c - c 4

18 Exmple 4: Qudrilterl with Perpendiculr igonls Given two sides, the lengths of the digonls nd the fct tht they re perpendiculr, wht re the lengths of the other two sides of qudrilterl? d +d +c+d +c-d -c+d -+c+d - c +c +c+d +c-d -c+d -+c+d -

19 Exmple 43: Intersection of ommon Tngent with xis of Symmetry of Two ircles Line hs length. is perpendiculr distnce x from line, nd is perpendiculr distnce y from. Find the distnce of the intersection point etween nd from nd : y x+y y E x x x+y

20 Wht if we chnge the digrm slightly so tht E is externl to : y x-y x x-y E y x

21 Exmple 44: Slope of the ngle isector Wht is the slope of the ngle isector of line with slope 0 nd line with slope m? m m m 0

22 Exmple 45: Loction of intersection of common tngents ircles nd hve rdii r nd s respectively. If the centers of the circles re prt, nd E is the intersection of the interior common tngent with the line joining the two centers, wht re the lengths E nd E? r r r+s E s r+s s

23 How out the exterior common tngent? r r -r+s s - s -r+s E

24 Exmple 46: ltitude of yclic Trpezium defined y common tngents of circles Given circles rdii r nd s nd distnce prt, wht is the ltitude of the trpezium formed y joining the intersections of the 4 common tngents with one of the circles? r J r s H s F G I E Notice tht this is symmetricl in r nd s, nd hence the trpezium in circle hs the sme ltitude.

25 Exmple 47: res yclic Trpezi defined y common tngents of circles Look t the rtio of the res of the trpezi in the previous exmple: z 0 r s -r - r s-s + r s -r + r s-s r J H s F G I E z 0 z 1 r s z 1 r s -r - r s-s + r s -r + r s-s

26 Exmple 48: Tringle formed y the intersection of the interior common tngents of three circles Notice tht if is the re of the tringle formed y the centers of the circles, then re STU is: ) )( )( ( t r t s s r rst P G M R T S Q J L H N E F U K O I r s t ++c +-c -+c -++c (r+s) (r+t) (s+t) c r t s

27 Exmple 49: istnce etween sides of rhomus Given rhomus with side length nd digonl, wht is the perpendiculr distnce etween opposite sides? + -

28 Exmple 50: ngles of Specific Tringles Here re some tringles, with their ngles displyed π 3 π 3 π 3

29 π 4 π 4 rctn rctn

30 Exmple 51: Sides of Specific Tringles Here re some specific ngle-defined tringles: π 4 3 π 6

31 π π 6 + 3

32 Exmple 5: 3 S I E S E F I N E ngles in the generl tringle c rccos - + +c c T R I N G L E W I T H S I E S N I N L U E N G L E c π+rctn - sin() -c+ cos()

33 T R I N G L E W I T H S I E S N N O N - I N L U E N G L E c π-rcsin sin() c T R I N G L E W I T H T W O N G L E S Here s the exterior ngle: s+t s t

34 Exmple 53: Some implied right ngles The medin of n isosceles tringle is lso its perpendiculr isector, nd ltitude: π

35 tringle whose medin is the sme length s hlf its se is right ngled: π

36 The digonls of kite re perpendiculr π

37 The line joining the intersection points of two circles is perpendiculr to the line joining their centers: π

38 Here is prticulr tringle (from the ook The urious Incident of the og in the Night ) -1+n π n 1+n

39 Exmple 54: Tringle defined y ngles nd side sin(t) sin(s+t) π-s-t sin(s) sin(s+t) s t

40 Exmple 55: ngle Tringle defined y two sides nd the included +c - c cos() c rctn - sin() -c+ cos()

41 Exmple 56: ngle Tringle defined y sides nd the non-included -rctn - - -c sin() -c cos() sin() -c+ -c sin() cos()+c cos() c π-rcsin c sin()

42 Exmple 57: Incenter The incenter is the intersection of the ngle isectors. We hve tringle with ngles nd nd se d: d cos(+) sin(+) - d cos(-+) sin(+) d (cos(+)-cos(-+)) sin(+) d We see tht the perpendiculr distnce to the other sides is the sme. This shows tht is the center of circle tngent to nd.

43 Exmple 58: Qudrilterl Formed y Joining the Midpoints of the sides of Qudrilterl -x x y 0 y + x 1,y 1 F x,y -x x y 1 y + 3 E x x y 1 y - 3 G x 0,y 0 H x x y 0 y - x 3,y 3 Quick inspection of the side lengths shows tht the new figure is prllelogrm

44 Exmple 59: Some mesurements on the Pythgors igrm rw right tringle nd sutend squre on ech side. The two red lines in the digrm re equl in length. G H +(--) F I +(--) E

45 Exmple 60: n unexpected tringle from Pythgors-like digrm Regrdless of the originl tringle the resulting tringle from this digrm is right ngledisosceles: +c + ++c +-c -+c -++c J F E G π c K H I Exmintion of the length J shows tht it is symmetric in nd, nd hence identicl to K.

46 Exmple 61: Theorem on Qudrilterls This theorem sttes tht if you drw squre on ech side of qudrilterl, then connect the center of opposite sides, the resulting lines hve the sme length, nd re perpendiculr. Here is the result in Geometry Expressions - + c + d -c d + e + e +c e +d e +e 4 + e +d+e +d-e -d+e -+d+e + e +c+e +c-e -c+e -+c+e + +d+e +d-e -d+e -+d+e +c+e +c-e -c+e -+c+e e J I H E e G c c K F d d L - + c + d -c d + e + e +c e +d e -e 4 + +d+e +d-e -d+e -+d+e +c+e +c-e -c+e -+c+e e If we crete the length of the other side we cn y creful exmintion see tht the lengths re identicl. lterntively, we cn do some simplifiction. Our constrints re necessrily symetric Geometry Expressions will not let you over-constrin the digrm, nd one digonl is sufficient to define the qudrlterl. However, we might expect the formul to e simpler if expressed in terms of oth digonls. lose inspection of the formul for the length shows tht it incorportes the squre of the other digonl of the figure, s well s Heron s formul for the res of the tringles nd. The following Mthemtic worksheet contins the formuls from Geometry Expressions for L the length of the desired line, nd f the length of the other digonl. little mnipultion gives simple formul for L^-f^/. This cn e simplified further y noting tht the remining terms re e^/ nd twice the re of the qudrlterl:

47 J-J- + c + d - c d + e + e + c e + d e + e 4 + e " + d + e " + d - e " - d + e " # - + d + e + e " + c+ e " + c- e " # - c + e " # - + c + e +" # + d + e " # + d - e " # - d + e " # - + d + e " # + c+ e " # + c - e " # - c + e " # - + c + en 1 e I,I- + c + d - c d + e + e +c e +d e +!! 4 +c- e e - c! - +c +e! +c! +! - +d +e! +d +! +c - e! +d - e! - c +e! - +c+e! +e! - d +e! - +d +e! +d +em J"# -J- + c + d - c d + e + e + c e + d e - e +" # 4 + d + e " # + d - e " # - d + e " # - + d + e " # + c+ e " # + c - e " # - c+ e " # - + c+ en 1! ei,i- + c + d - c d + e + e +c e + d e - e +! 4 +c - e! +d - e! - c +e! - +c+e! +c+e! - d +e! - +d +e! +d +em 1 4e I- + c + d - c d + e + e +c e +d e +e +!! 4 +c - e e - c +e! - +c+e! +!! +d - e e - d +e! - +d +e! +d +! +c - e! +d - e! - c +e! - +c+e! +c+e! - d +e! 1 e I- + c + d - c d + e + e +c e +d e - e +! 4 +c - e! +d- e! - c +e! - +c +e! +c +e! - d +e! - +d +e! +d+em +! 1 Ie +c- e! - c +e! - +c +e! +c +! +d - e! - d +e! - +d +e! +d +em In[6]:= L = 1 e Out[6]= In[4]:= f = 1 e Out[4]= In[7]:= L^ Out[7]= In[8]:= f^ Out[8]= In[11]:= Simplify@L^ - f^ Out[11]= e f L = + + From which we cn derive tht:! +d - e e - d +e - +d +e! +d +em

48 res re simpler when expressed in terms of ngles. Here is revision of the digrm with ngles inserted. This gives us more of clue of how to prove the result: +c + d + c sin(θ)+c d sin(φ)- d sin(θ+φ)- c cos(θ)-c d cos(φ) J + c + c sin(θ) c +d - c d cos(φ) π 4 H θ π 4 G I E +c - c cos(θ) F L d π φ 4 c c + d +c d sin(φ) K

49 Exmple 6: Rectngle ircumscriing n Equilterl Tringle F E irectly from the digrm we hve the following theorem: The re of the lrger right tringle is the sum of the res of the smller two. This ppers in pge 19-1 of Mthemticl Gems, y Ross Honserger (nd vrious other plces).