Jarak Dimensi

Setelah menyaksikan tayangan ini anda dapat Menentukan jarak antara unsur-unsur dalam ruang dimensi tiga

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],A B Jarak dua titik

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],a cm a cm a cm P A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],A B C D H E F G a cm a cm a cm

a cm P ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],A B C D H E F G

[object Object],A g Jarak titik dan garis Peragaan ini, menunjukan jarak titik A ke garis g adalah panjang ruas garis yang ditarik dari titik A dan tegak lurus garis g

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],5 cm 5 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],5 cm 5 cm AH = (AH diagonal sisi) AH = Jadi jarak A ke HG = 5 √2 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],6 cm 6 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],6 √2 cm 6 cm P 6 √3 cm ? A B C D H E F G A B G P 6 √3 6 6 √2

Lihat segitiga ABG Sin  A = = = BP = BP = 2√6 ? Jadi jarak B ke AG = 2 √6 cm 2 A B G P 6 √3 6 6 √2

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],12 cm 12 √2 cm T C A B D

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],P 12 √2 6 √2 6 √2 12 cm 12 √2 cm T C A B D

[object Object],[object Object],[object Object],[object Object],[object Object],6 cm 6 cm Titik P pada pertengahan FG. Jarak titik A dan garis DP adalah…. P A B C D H E F G

6 cm 6 cm P ,[object Object],[object Object],Q 6 √2 cm R DP = = = A B C D H E F G P A D G F 6 cm 3 cm

[object Object],DP = Luas segitiga ADP ½DP.AQ = ½DA.PR 9.AQ = 6.6√2 AQ = 4√2 Jadi jarak A ke DP = 4√2 cm 4 Q 6 √2 cm R P A D G F 6 cm 3 cm

[object Object],Garis tegak lurus sebuah bidang jika garis tersebut tegak lurus dua buah garis berpo- tongan yang ter- dapat pada bidang  g a b g  a, g  b, Jadi g  V V

[object Object],Peragaan ini menunjukan jarak antara titik A ke bidang V adalah panjang ruas garis yang menghubungkan tegak lurus titik A ke bidang V A  V

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],10 cm P A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],10 cm P Jadi jarak A ke BDHF = 5 √2 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],8 cm 12 cm T C A B D

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],8 cm 12 cm P T C A B D

AP = ½ AC = 4√2 TP = = = = = 4√7 Jadi jarak T ke ABCD = 4 √7 cm 8 cm T C A B D 12 cm P

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],9 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],9 cm P T CP = ⅓CE = ⅓.9√3 = 3√3 Jadi jarak C ke BDG = 3√3 cm A B C D H E F G

[object Object],Peragaan menunjukan jarak antara garis g ke garis h adalah panjang ruas garis yang menghubungkan tegak lurus kedua garis tersebut P Q g h

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],4 cm ,[object Object],[object Object],[object Object],A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],4 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],4 cm P Q A B C D H E F G

[object Object],Peragaan menunjukan Jarak antara garis g ke bidang V adalah panjang ruas garis yang menghubungkan tegak lurus garis dan bidang g g V

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],8 cm P A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],8 cm P Jadi jarak A ke BDHF = 4 √2 cm A B C D H E F G

[object Object],peragaan, menunjukan jarak antara bidang W dengan bidang V adalah panjang ruas garis yang tegak lurus bidang W dan tegak lurus bidang V Jarak Dua Bidang V W W

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],6 cm 6 cm P Q Jadi jarak AFH ke BDG = 4 √2 cm A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],12 cm Titik K, L dan M berturut-turut merupakan titik tengah BC, CD dan CG. Jarak antara bidang AFH dan KLM adalah…. K L M A B C D H E F G

[object Object],[object Object],[object Object],[object Object],[object Object],12 cm Sehingga jarak E ke AFH = ⅓EC =⅓.12√3 = 4√3 Berarti jarak BDG ke C juga 4√3 L A B C D H E F G

12 cm BDG ke C juga 4√3 Jarak BDG ke KLM = jarak KLM ke C = ½.4√3 = 2√3 K L M Jadi jarak AFH ke KLM = jarak AFH ke BDG + jarak BDG ke KLM = 4√3 + 2√3 = 6√3 cm A B C D H E F G

Jarak Dimensi

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Jarak Dimensi