Jawaban:
Penjelasan dengan langkah-langkah:
a. Nilai P
AB = AC
(2p-4)=(p+3)
(2p-p)=3+4
(P)=7 [✓]
b. Panjang sisi segitiga ABC
AB = (2p-4)
= (2(7)-4)
= (14-4)
= (10) cm [✓]
AC = (p+3)
= (7+3)
= 10 cm [✓]
BC = 15 cm [✓]
c. Keliling segitiga ABC
AB + AC + BC
(10 cm + 10 cm +15 cm)
(35 cm) [✓]
d. Luas segitiga ABC
L = √s(s-a)(s-b)(s-c)
= √17,5(17,5-10)(17,5-10)(17,5-15)
= √17,5(7,5)(7,5)(2,5)
= √2.460,9375
= 49,6078 cm² [✓]
s = 1/2 (a+b+c)
= 1/2 (10+10+15)
= 1/2 (35)
= 17,5 cm
Detail Jawaban:
Mapel : Matematika
Kelas : 11 / XI SMA
Materi : Trigonometri (Segitiga)
Kode Kategorisasi : -
Kata Kunci : Trigonometri (Segitiga)
Demikian
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Jawaban:
Trigonometri
Segitiga
Penjelasan dengan langkah-langkah:
a. Nilai P
AB = AC
(2p-4)=(p+3)
(2p-p)=3+4
(P)=7 [✓]
b. Panjang sisi segitiga ABC
AB = (2p-4)
= (2(7)-4)
= (14-4)
= (10) cm [✓]
AC = (p+3)
= (7+3)
= 10 cm [✓]
BC = 15 cm [✓]
c. Keliling segitiga ABC
AB + AC + BC
(10 cm + 10 cm +15 cm)
(35 cm) [✓]
d. Luas segitiga ABC
L = √s(s-a)(s-b)(s-c)
= √17,5(17,5-10)(17,5-10)(17,5-15)
= √17,5(7,5)(7,5)(2,5)
= √2.460,9375
= 49,6078 cm² [✓]
s = 1/2 (a+b+c)
= 1/2 (10+10+15)
= 1/2 (35)
= 17,5 cm
Detail Jawaban:
Mapel : Matematika
Kelas : 11 / XI SMA
Materi : Trigonometri (Segitiga)
Kode Kategorisasi : -
Kata Kunci : Trigonometri (Segitiga)
Demikian
Semoga bermanfaat dan bermanfaat!