Rectangles, Rhombuses, and Squares Kelompok 3 Rectangles, Rhombuses, and Squares Annisa Luthfi Fadhilah Ma’ruf ; Rosyida Khikmawati ; Rizqi Dwi Maharani ; Nadiatul Khikmah

Theorem 8-9 A parallelogram is a rectangle if an only if its diagonal are congruent Jajargenjang adalah sebuah persegi panjang jika dan hanya jika digonalnya kongruen

We must prove two things 1. if the diagonals of a parallogram are congruent, then the paralellogram is rectangle 2. if a parallelogram is a rectangle, then the diagonals are congruent

Jadi, segitiga ABD dan segitiga BAC kongruen (SSS postulate) Pernyataan Alasan AC AB Diketahui AB AB Berhimpit AD BC Definisi jajargenjang Jadi, segitiga ABD dan segitiga BAC kongruen (SSS postulate)

Theorem 8-10 A parallelogram is a rhombus if an only if its diagonal are perpendicular to each other Jajargenjang adalah belah ketupat jika dan hanya jika diagonalnya tegak lurus dengan diagonal yang lainnya

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Theorem 8-11 A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles.

PROOF Given: Prove that parallelogram ABCD would be rhombuses 30 60 D C 90 O

Answer: ˂O = 90 ˂ DAO ˂ BAO ˂ BCO ˂ DCO ˂CDO ˂ADO ˂ ABO ˂ CBO So, the parallelogram would be rhombuses A B 30 60 D C 90 O

Theorem 8-12 The segment joining the midpoints of the two nonparallel sides of a trapezoid is parallel to the two bases and has a length equal to one half the sum of the lengths of the bases.

PROOF Given: ABCD is a trapezoid with DC AB E the midpoint of AD and F the midpoint of BC Prove: EF AB, EF DC and EF= ½ (AB+CD)

Extend AB and DF to meet at G Extend AB and DF to meet at G. Then prove that F is midpoint of DG and use the Midsegment Theorem A B C D G F E

It show that EF AB, EF DC and EF= ½ ( AB+CD) Statement Reasons 1. Extend AB Construction 2. Draw DF Intersecting AB at G 3. DC AB Definition of trapezoid 4. ˂BGF ˂DFC Alternate interior angles 5. CF BF F midpoint of BC 6. ˂BFG ˂DFC Sudut bertolak belakang 7. ∆BFG ∆CFD ASA postulate 8. DF GF CPCTC 9. F is midpoint of DG Statement 8 10. EF AB and EF DC Midsegment Theorem It show that EF AB, EF DC and EF= ½ ( AB+CD)

An isosceles trapezoid is a trapezoid with congruent nonparallel sides B A AD and BC are nonparallel sides. <A and <B are called base angles. <C and <D together are another pair of base angles.

Theorem 8-13 In an isosceles trapezoid base angles are congruent and the diagonals are congruent

13. Given : ABCD is an isosceles trapezoid with AB ││CD Prove : AC BD

∆ Plan : Draw perpendiculars from D to AB intersecting at E and from C to AB intersecting at F and prove ∆ DEA ∆ CFB

˂ Statements Reason Definition an isosceles trapezoid DA CB Definition an isosceles trapezoid EA BF Definition of perpendicular ∆ DEA ∆ CFB HL theorem AB BA Reflexive DAB ˂ CBA Congruent sup ∆ DAB ∆ CBA SAS Postulate BD AC CPCTC ˂

15. Given : ABCD is an isosceles trapezoid with AB││ CD Prove : <A <B Plan : Draw DE ││CB with E on AB

Statements Reason DEBC is parallelogram DE ││CB DE CB Theorem 8-2 Parallelogram DE DA Definition of an isosceles triangle AB BA Reflexive ˂ A ˂DEA Definition of isosceles triangle ˂DEA ˂B Corresponding angle ˂ A ˂B Transitive

Theorem 8-14 The sum of the measure of the angles of a convex polygon of n sides is (n-2) 180

Masih ingat dengan rumus yang satu ini?? α1+ α2+ α3+… + αn = (n-2) 180 dengan   α1,α2,α3,… ,αn adalah besar sudut-sudut dalam dari suatu bangun datar segi-n.

Segitiga Gambarlah sebuah segitiga, kemudian bagi segitiga tersebut menjadi segitiga (tidak harus sama besar) dan berikan label pada sudut-sudut yang terbentuk seperti gambar berikut

Θ1+ Θ2+ Θ3 = 3x180 – (ß1+ ß2+ ß3+ ß4+ ß5+ ß6) Dari gambar di atas kita peroleh tiga persaman berikut : Θ1 = 180-(ß1+ ß2) Θ2 = 180-(ß3+ ß4) Θ3 = 180-(ß5+ ß6) Jumlahkan ke tiga  persamaan tersebut sehingga diperoleh: Θ1+ Θ2+ Θ3 = 3x180 – (ß1+ ß2+ ß3+ ß4+ ß5+ ß6) 360 = 3x180 – (ß1+ ß2+ ß3+ ß4+ ß5+ ß6) ß1+ ß2+ ß3+ ß4+ ß5+ ß6 = (3-2)x180

Segiempat Gambarlah sebuah segiempat, kemudian bagi segiempat tersebut menjadi 4 segitiga (tidak harus sama besar) dan berikan label pada sudut-sudut yang terbentuk seperti gambar berikut

Dari gambar di atas kita peroleh tiga persaman berikut : Θ1 = 180-(ß1+ ß2) Θ2 = 180-(ß3+ ß4) Θ3 = 180-(ß5+ ß6) Θ4 = 180-(ß7+ ß8)

Jumlahkan ke empat persamaan tersebut sehingga diperoleh: Θ1+ Θ2+ Θ3+ Θ4 = 4x180 – (ß1+ ß2+ ß3+ ß4+ ß5+ ß6 + ß7+ ß8) 360 = 4x180 – (ß1+ ß2+ ß3+ ß4+ ß5+ ß6 + ß7+ ß8) ß1+ ß2+ ß3+ ß4+ ß5+ ß6 + ß7+ ß8 = (4-2)x180

Theorem 8-15    

Theorem 8-16 The sum of the measures of the exterior angles a polygon one at each vertex, is 360 Bukti Jumlah n buah sudut dalam & sudut luar = n.180° Jumlah n buah sudut dalam =(n-2)180°- Jumlah n buah sudut luar =2.180°

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Sum of the measures of the angles Polygon Number of sides Number of triangles Sum of the measures of the angles Quadrilateral Pentagon Hexagon . n-gon 4 5 6 n 2 3 n-2

Sum of the measures of the angles Sum of the measures of the angles Polygon Number of sides Number of triangles Sum of the measures of the angles Quadrilateral Pentagon Hexagon . n-gon 4 5 6 n 2 3 n-2 Polygon Number of sides Number of triangles Sum of the measures of the angles Quadrilateral Pentagon Hexagon . n-gon 4 5 6 n 2 3 n-2