Jawab:
Penjelasan dengan langkah-langkah:
a] s = 1/2 . (AB + BC + AC)
s = 1/2 . (12 + 15 + 17)
s = 1/2 . 44
s = 22 cm
Luas segitiga
= √(s . (s - a) . (s - b) . (s - c))
= √(22 . (22 - 12) . (22 - 15) . (22 - 17))
= √(22 . 10 . 7 . 5)
= √(2 . 11 . 2 . 5 . 7 . 5)
= 10 √77 cm²
r = AB . BC . AC / (4 . luas segitiga)
r = ((12 . 15 . 17 / (4 . 10 . √77)) . ((√77)/(√77))
r = (153/154) . √77 cm
b] Luas daerah yang diarsir
= luas lingkaran - luas segitiga
= π . r . r - 10 √77
= 22/7 . (153/154) . √77 . (153/154) . √77 - 10 √77
= 238,87 - 87,75
= 151,12 cm²
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Jawab:
Penjelasan dengan langkah-langkah:
a] s = 1/2 . (AB + BC + AC)
s = 1/2 . (12 + 15 + 17)
s = 1/2 . 44
s = 22 cm
Luas segitiga
= √(s . (s - a) . (s - b) . (s - c))
= √(22 . (22 - 12) . (22 - 15) . (22 - 17))
= √(22 . 10 . 7 . 5)
= √(2 . 11 . 2 . 5 . 7 . 5)
= 10 √77 cm²
r = AB . BC . AC / (4 . luas segitiga)
r = ((12 . 15 . 17 / (4 . 10 . √77)) . ((√77)/(√77))
r = (153/154) . √77 cm
b] Luas daerah yang diarsir
= luas lingkaran - luas segitiga
= π . r . r - 10 √77
= 22/7 . (153/154) . √77 . (153/154) . √77 - 10 √77
= 238,87 - 87,75
= 151,12 cm²