Kelompok 3 Rectangles Rhombuses and Squares Annisa Luthfi

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

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

/ 2 1 1 1 2 2 2 1

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

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

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

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 E the midpoint of AD and F the midpoint of BC Prove: EF AB, EF DC and EF= ½ (AB+CD) AB

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

Statement Reasons 1. Extend AB 2. Draw DF Intersecting AB Construction at G 3. DC AB 4. ˂BGF 5. CF 6. ˂BFG ˂DFC Sudut bertolak belakang 7. ∆BFG ∆CFD ASA postulate 8. DF 9. 10. Construction Definition of trapezoid Alternate interior angles ˂DFC BF F midpoint of BC GF CPCTC F is midpoint of DG EF AB and EF Statement 8 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 D A C B 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 A

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

Statements Reason DA CB Definition an isosceles trapezoid EA BF Definition of perpendicular ∆ DEA AB ∆ CFB BA HL theorem 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 ˂A ˂B ˂B Corresponding angle 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? ? /α 1+ α 2+ α 3+… + αn = (n-2) 180 /dengan α 1, α 2, α 3, … , αn adalah besar sudut dalam dari suatu bangun datar segi-n.

Segitig a Gambarlah sebuah segitiga, kemudian bagi segitiga tersebut menjadi 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 : / Θ 1 = 180 -(ß 1+ ß 2) / Θ 2 = 180 -(ß 3+ ß 4) / Θ 3 = 180 -(ß 5+ ß 6) / Jumlahkan ke tiga persamaan tersebut sehingga diperoleh: Θ 1+ Θ 2+ Θ 3 = 3 x 180 – (ß 1+ ß 2+ ß 3+ ß 4+ ß 5+ ß 6) 360 = 3 x 180 – (ß 1+ ß 2+ ß 3+ ß 4+ ß 5+ ß 6) ß 1+ ß 2+ ß 3+ ß 4+ ß 5+ ß 6 = (3 -2)x 180

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 : Θ 1 = 180 -(ß 1+ ß 2) Θ 2 = 180 -(ß 3+ ß 4) Θ 3 = 180 -(ß 5+ ß 6) Θ 4 = 180 -(ß 7+ ß 8)

/ Jumlahkan ke empat persamaan tersebut sehingga diperoleh: Θ 1+ Θ 2+ Θ 3+ Θ 4 = 4 x 180 – (ß 1+ ß 2+ ß 3+ ß 4+ ß 5+ ß 6 + ß 7+ ß 8) 360 = 4 x 180 – (ß 1+ ß 2+ ß 3+ ß 4+ ß 5+ ß 6 + ß 7+ ß 8) ß 1+ ß 2+ ß 3+ ß 4+ ß 5+ ß 6 + ß 7+ ß 8 = (4 -2)x 180

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 Jumlah n buah sudut dalam Jumlah n buah sudut luar = n. 180° =(n-2)180°=2. 180°

3 2 4 5 1 2 3 1 4 5

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

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