Question

Mapel: Matematika
Materi: Kekongruenan dan Kesebangunan
Kelas: 9 SMP
Bantu jawab tinggal 2 soal gk ngerti caranya (╥﹏╥)​

Answer 1

Materi : Kesebangunan dan Kekonggruenan

Soal Nomor 6

UP : UQ = 3 : 2 , PQ = 5

Panjang UT

a¹ = PR = 15 , b¹ = PQ = 5

a² = UT , b² = UQ = 2

___________________/

a²/a¹ = b²/b¹

UT/PR = UQ/PQ

UT/15 = ⅖

UT = 15 × ⅖

UT = 3 × 2

UT = 6 cm

Maka Panjang ST = PR - UT = 15 - 6 = 9 cm

Soal Nomor 7

a¹ = 4 , b¹ = 12 , c¹ = 8

a² = BC = CE + 8 , b² = 14 , c² = AB = AD + 12

__________________________

Panjang CE

a²/a¹ = b²/b¹

( CE + 8 )/4 = 14/12

6( CE + 8 ) = 28

3( CE + 8 ) = 14

CE + 8 = 4⅔

CE = - 3⅓ cm

Panjang AD

c²/c¹ = b²/b¹

( AD + 12 )/8 = 14/12

6( AD + 12 ) = 56

AD + 12 = 9⅓

AD = - 2⅔ cm

Semoga bisa membantu

[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]

Answer 2

Nomor 6

[tex]QU =\frac{2}{3}UP\\\\\frac{QU}{UP} = \frac{2}{3} \: \: \: \: \: \: \: [/tex]

..

[tex]\begin{aligned}\frac{QU}{QP} &= \frac{UT}{PR}\\\\\frac{2}{2+3} &= \frac{UT}{15} \\\\\frac{2}{5} &= \frac{UT}{15} \\\\5 \times UT&=15 \times 2 \\\\UT&= \frac{30}{5} \\ \\UT&=6\end{aligned}[/tex]

..

Panjang TS

= PR - UT

= 15 - 6

= 9

[tex] \\ [/tex]

Nomor 7

Terlampir

#LearnWithPanda