equilateral regular irregular

Transcription

1 polygon three polygon side common sides vertex nonconsecutive diagonal triangle quadrilateral pentagon hexagon heptagon octagon nonagon decagon dodecagon n-gon equiangular concave exterior equilateral regular irregular diagonal 1

2 diagonals convex regular exterior diagonals interior equals convex triangles sides convex sum two exterior 2

3 polygon hexagon polygon heptagon not a polygon not a polygon polygon nonagon not a polygon irregular convex irregular concave regular convex regular convex irregular concave irregular concave 3

4 (n 2)180 (7 2)180 (5) (n 2)180 (10 2)180 (8) =144 (5 2)180 =540 m A+m B+m C+m D+m E=540 35c+18c+32c+32c+18c= c=540 m A=35(4)=140 m B=m E=18(4)=72 m C=m D=32(4)= c=4 n( ) (n 2)180 =135 n (n 2)180=135n 180n 360=135n 180n 180n 360= 45n n=8 A dodecagon has 12 sides. sum of ext. s= =30 15b+18b+33b+16b+10b+28b= b= b=3 4

5 four quadrilateral special quadrilaterals two parallel parallelogram sides side 5

6 CF=74 m DEF=42 DF=2(31)=62 YZ=XW 8a 4=6a+10 6a 6a 2a 4= a= a=7 YZ=8a 4 =8(7) 4=52 m W+m Z=180 18b 11+9b+2=180 27b 9= b= b=7 m Z=9b+2 =9(7)+2=65 EJ=JG 3w=w+8 w w 2w=8 2 2 w=4 JG=w+8 =4+8=12 HJ=JF 2z=4z 9 4z 4z 2z= z=4.5 FH=2(2z) =2(2(4.5))=18 K( 2,2) L(2,4) J(3, 4) M(7, 2) Opposite sides are parallel Find slope of KL by counting Over 4 up 2 From point J, go over 4 and up 2, this is your point M. M(7, 2) 6

7 properties definition parallelogram conditions 7

8 opposite parallel both JK=15a 11 =15(3) 11=34 LM=10a+4 m Q=(6y+7) =10(3)+4=34 =(6(6.5)+7) =46 KL=5b+6 =5(9)+6=51 JK=LM KL=JM JM=8b 21 =8(9) 21=51 Since both pairs of opposite sides are congruent, JKLM is a parallelogram. m S=(8y 6) =(8(6.5) 6) =46 m R=(15x 16) =(15(10) 16) = =180 Since one is supplementary to its consecutive s, PQRS is a parallelogram. 8

9 One angle is supplementary to both its consecutive angles. The quadrilateral is a parallelogram. One pair of opposite angles are congruent. This is not enough information. Both pairs of opposite angles are congruent. The quadrilateral is a parallelogram. No. Two pairs of consecutive sides congruent does not form a parallelogram. JK 1 ( 6) 5 m= = 4 ( 1) 3 LM m= = AB m= = CD m= = KL 5 ( 1) 6 3 m= = = 4 ( 4) 8 4 JM 0 ( 6) 6 3 m= = = 7 ( 1) 8 4 d= (6 2) 2 +(2 3) 2 = 17 d= (1 5) 2 +(1 0) 2 = 17 JK LM KL JM The quadrilateral has two pairs of parallel sides, therefore it is a parallelogram. AB CD AB=CD The quadrilateral has one pair of parallel and congruent sides, therefore it is a parallelogram. 9

10 right rectangle properties parallelogram rhombus congruent special 10

11 rectangle rhombus properties parallelograms square right congruent parallelogram rectangle rhombus special 11

12 58 35 TR=35 CE=1/2(58)= HM=1/2(86)=43 HJ=48 in. HK=JG =2(30.8)=61.6 in. RS=QR 3a=4a 14 4a 4a a= a=14 QP=RS =3(14)=42 m QRP=1/2(180 78) =51 WV=XT 13b 9=3b+4 3b 3b 10b 9= b= b=1.3 TV=XT =3(1.3)+4 =7.9 m TZV=90 14a+20= a= a=5 m VTZ=m XTZ =(5(5) 5) =20 CG=GF 5a=3a+17 3a 3a 2a= a=8.5 CD=CG =5(8.5) =42.5 m GCD+m CDF=180 b+3+6b 40=180 7b 37= b= b=31 m GCH=1/2m GCD =1/2(31+3) =17 12

13 First we need to find the slopes of the diagonals. EG FH 0 ( 1) m= = m= = 3 ( 4) 7 0 ( 1) 1 Therefore, EG FH. Then we need to find the lengths of the diagonals. d= (3 ( 4)) 2 +(0 ( 1)) 2 = 50 d= (0 ( 1)) 2 +( 4 3) 2 = 50 Therefore, EG=FH. =( ) =( ) Then we need to find the midpoints of the diagonals ( 4) 1 1 ( 2, 2 ) 2, 2 ( 2, 2 ) 2, 2 The diagonals are congruent, perpendicular, and have the same midpoint. They are congruent perpendicular bisectors of each other. First we need to find the slopes of the diagonals. SV TW 3 ( 4) m= = m= = 6 ( 5) Therefore, SV TW. Then we need to find the lengths of the diagonals. d= (6 ( 5)) 2 +( 3 ( 4)) 2 = 122 d= (1 0) 2 +( 9 2) 2 = 122 =( ) Therefore, SV=TW. =( ) Then we need to find the midpoints of the diagonals ( 2, 2 ) 2, 2 ( 2, 2 ) 2, 2 The diagonals are congruent, perpendicular, and have the same midpoint. They are congruent perpendicular bisectors of each other. 13

14 parallelogram rectangle rhombus rectangle rhombus square 14

15 P( 1,4) S(1,1) Q(2,6) R(4,3) Find the lengths of the diagonals. PR= (4 ( 1)) 2 +(3 4) 2 = 26 SQ= (2 1) 2 +(6 1) 2 = 26 The diagonals are congruent. Parallelogram PQRS is a rectangle. PR=SQ Find the slopes of the diagonals. PR m= = 4 ( 1) 5 SQ m= = The diagonals are perpendicular. Parallelogram PQRS is a rhombus. PR SQ Since parallelogram PQRS is both a rhombus and a rectangle, it is a square. Find the lengths of the diagonals. W(0,1) X(4,2) WY= (3 0) 2 +( 2 1) 2 = 18 ZX= (4 ( 1)) 2 +(2 ( 3)) 2 = 50 WY ZX Z( 1, 3) Y(3, 2) The diagonals are not congruent. Parallelogram WXYZ is not a rectangle. Find the slopes of the diagonals. WY m= = = ZX 2 ( 3) 5 m= = =1 4 ( 1) 5 The diagonals are perpendicular. Parallelogram WXYZ is a rhombus. WY ZX Since WXYZ is not a rectangle, it is not a square. 15

16 kite quadrilateral two trapezoid one base nonparallel legs angles consecutive base 16

17 legs congruent isosceles isosceles midsegment endpoints midpoints Triangle Trapezoid 17

18 m BCD=2(90 52) =76 m ABC=(90 27) +52 =115 CD= m QPS=(90 39) +59 =110 m PSR=2(90 59) =62 QP=15 m A= m F= =80 =131 AD=BC 12x 11=9x 2 9x 9x 3x 11= x=9 3 3 x=3 m S=m P 2a 2 54=a a 2 a 2 a 2 54= a 2 =81 a 2 = 81 a= = EH= =8 18

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