PT = 12 cm
PQ = 9 + 11 = 20 cm
PR = 12 + x
TR = x = ? cm
→ [tex] \tt \frac{tp}{pq} = \frac{ps}{PR} [/tex]
→ [tex] \tt \frac{12}{20} = \frac{9}{12 + x} [/tex]
→ 20(9) = 12(12 + x)
→ 180 = 144 + 12x
→ -12x = 144 - 180
→ -12x = -36 cm
→ x = 3 cm
Maka, panjang TR adalah B.) 3 cm.
Mapel : 2 - Matematika
Kelas : 7 - SMP
Bab : 3 - Garis dan sudut
Kode kategorisasi : 7.2.3
Kata kunci : Panjang garis, Segitiga
[tex]\purple{\boxed{\blue{\boxed{\green{\star{\orange{\ \: \: \mathcal{JK} \: \: \: {\green{\star}}}}}}}}} [/tex]
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Verified answer
Diketahui :
PT = 12 cm
PQ = 9 + 11 = 20 cm
PR = 12 + x
TR = x = ? cm
Penyelesaian :
→ [tex] \tt \frac{tp}{pq} = \frac{ps}{PR} [/tex]
→ [tex] \tt \frac{12}{20} = \frac{9}{12 + x} [/tex]
→ 20(9) = 12(12 + x)
→ 180 = 144 + 12x
→ -12x = 144 - 180
→ -12x = -36 cm
→ x = 3 cm
Kesimpulan :
Maka, panjang TR adalah B.) 3 cm.
Detail Jawaban.•:。
Mapel : 2 - Matematika
Kelas : 7 - SMP
Bab : 3 - Garis dan sudut
Kode kategorisasi : 7.2.3
Kata kunci : Panjang garis, Segitiga
[tex]\purple{\boxed{\blue{\boxed{\green{\star{\orange{\ \: \: \mathcal{JK} \: \: \: {\green{\star}}}}}}}}} [/tex]