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Verified answer
No. 1cari segitiga yang sisi-sisinya membentuk tripel Phytagoras.
→ (1)
16² + 30²
= 256 + 900
= 1156
= 34²
→ (2)
21² + 72²
= 441 + 5184
= 5625
= 75²
→ (3)
11² + 60²
= 121 + 3600
= 3721
= 61²
No. 2
tripel phytagoras = (b) (c) & (e)
(b)
8² + 15²
= 64 + 225
= 289
= 17
(c)
10² + 24²
= 100 + 576
= 676
= 26²
(e)
20² + 21²
= 400 + 441
= 841
= 29²
No. 3
a.
AC² = AD² + CD²
AC² = 9² + 12²
AC² = 81 + 144
AC² = 225
AC = 15
b.
BC² = DB² + DC²
BC² = 16² + 12²
BC² = 256 + 144
BC² = 400
BC = 20
c.
segitiga ABC siku-siku di C , maka
AC² + BC² = AB²
15² + 20² = 25²
225 + 400 = 625 (terbukti)
No.4
gunakan rumus jarak antar 2 titik (x1,y1) dan (x2,y2)
jarak = √(x1 - x2)² + (y1 - y2)²
diketahui titik-titik
D(-2,-1)
E(2,2)
F(5,-2)
jarak DE = panjang DE
DE = √(-2 - 2)² + (-1 -2)²
DE = √(16 + 9)
DE = √25
DE = 5
jarak EF = panjang EF
EF = √(2 - 5)² + (2 + 2)²
EF = √(9 + 16)
EF = √25
EF = 5
jarak DF = panjang DF
DF = √(-2 - 5)² + (-1 + 2)²
DF = √(49 + 1)
DF = √50
DF = 5√2
apakah segitiga DEF siku-siku ??
periksa
5² + 5²
= 25 + 25
= 50
= (5√2)²
jawab : ya, segitiga DEF adalah segitiga siku-siku.