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MATEMATIKA
PERSAMAAN KUADRAT

NAMA KELOMPOK : A ( KLS X AK 1 )
1.
2.
3.
4.
5.

NI PUTU NIA ANDRIANI
NI LUH PUTU EKA ANGGRENI
NI PUTU AYU PURNAMA YANTI
NI MADE HERLIANA KRISNANDA
NI LUH GEDE TIA YUNIKA SARI

( 01 )
( 04 )
( 19 )
( 26 )
( 27 )

SMK N 1 TABANAN
TH. PEL. 2013/2014
p

1. Diket : p
= 60 m
l
= 30 m
Lrencana = 1000 m

Tanah kosong berukuran 60 m x 30 m

l

Dit

: persamaan kuadrat…?

Rumus :
( p – x )( l – x ) = Lrencana
Jawab :
( p – x )( l – x )
( 60 – x )(30 – x )
1800 – 60x – 30x + x2
1800 – 90x + x2
1800 – 1000 – 90x + x2
800 – 90x + x2
800 – 80x – 10x + x2
80 ( 10 – x ) – x ( 10 – x )
( 80 – x )(10 – x )

= Lrencana
= 1000
= 1000
= 1000
=0
=0
=0
=0
=0

x = 80 atau x = 10

Diuji :
Yang digunakan adalah x = 10
( p – x )( l – x )
60 – 10 . 30 – 10
50 . 20

2. Diket : ps
ls
t
pa
la

= 50 cm
= 40 cm
=x
=50 – 2x
= 40 – 2x

= Lrencana
= 1000
= 1000

x

ls

x

pp pa
la

ps
Dit

: Persamaan kuadrat…?
Volume tempat air/balok…?

Keterangan :
ps = panjang seng

Rumus : pa . la = La

ls

Jawab :
pa . la

= lebar seng

t

V =p.l.t

= tinggi

pa = panjang alas

= La

la = lebar alas

( 50 – 2x ) ( 40 – 2x )

= 200

2000 – 100x – 80x + 4x2

= 200

2000 – 180x + 4x2

= 200

2000 – 200 – 180x + 4x2

=0

1800 – 180x + 4x2

=0

450 – 45x + x2

=0

450 – 30x – 15x + x2

=0

30 ( 15 – x ) – x ( 15 – x )

=0

( 30 – x ) ( 15 – x )

=0

La

= luas alas

x = 30 atau x = 15

Diuji :
Yang digunakan adalah x = 15
pa

= 50 – 2x

la

= 40 – 2 x

= 50 – 2 . 15
= 50 – 30

= 40 – 30

= 20 cm
t

= 40 – 2 . 15

= 10 cm

=x
= 15 cm

V

=p .l .t
= 20 . 10 . 15
= 3000 cm3
3. Diket

: t = 3 cm

Dit

: jari-jari kerucut…?

Rumus

: V =
V1 =
V2 =

Jawab

t
( r + 242 ) 3
r2( 3 + 24 )

:

V1

=
( r + 242 ) . 3

( r + 242 ) . 3
3 ( r2+ 48r + 242 )
3r2 + 3 . 48r + 3 . 24 .24
r2 + 48r + 24 . 24
0
0
0
0
r = 12

=

V2
r2( 3 + 24 )

= r2 ( 3 + 24 )
= 27r2
= 27r2
= 9 r2
= 8 r2 - 48r - 24 . 24
= r2 - 6r - 24 . 3
= r2 - 6r - 72
= ( r – 12 ) ( r + 6 )

atau r = - 6

Jadi , jari – jari kerucut semula adalah 12

4. Diket :

Dit

:

p1 = x jam
P2 = x + 1 jam
t1 + t2 = 1,2 jam
t2 untuk mencetak 1 buku…?

Keterangan :
p1 : printer satu
p2 : printer dua
t1 : waktu printer satu
t2 : waktu printer dua
Jawab :
= 1,2

= 1,2

= 1,2

= 1,2
x2 + x

= 1,2 ( 2x + 1 )

x2 + x

= 2,4x + 1,2

x2 + x – 2,4x – 1,2

=0

x2 – 1,4x – 1,2

=0

10x2 – 14x – 12x

=0

10x2 – 20x + 6x – 12x = 0
10x ( x – 2 ) + 6 ( x – 2 ) = 0
( 10x + 6 )( x – 2 )

=0

10x + 6 = 0

x–2=0

10x

= -6

x

x

= -6 : 10

x

= - 0,6

Yang digunakan adalah x = 2
p1

= x jam
= 2 jam

=2
p2

= x jam + 1 jam
= 2 jam + 1 jam
= 3 jam

Jadi , waktu yang diperlukan printer ke-dua untuk mencetak satu set buku adaah 3 jam.

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Klmpk matik

  • 1. MATEMATIKA PERSAMAAN KUADRAT NAMA KELOMPOK : A ( KLS X AK 1 ) 1. 2. 3. 4. 5. NI PUTU NIA ANDRIANI NI LUH PUTU EKA ANGGRENI NI PUTU AYU PURNAMA YANTI NI MADE HERLIANA KRISNANDA NI LUH GEDE TIA YUNIKA SARI ( 01 ) ( 04 ) ( 19 ) ( 26 ) ( 27 ) SMK N 1 TABANAN TH. PEL. 2013/2014
  • 2. p 1. Diket : p = 60 m l = 30 m Lrencana = 1000 m Tanah kosong berukuran 60 m x 30 m l Dit : persamaan kuadrat…? Rumus : ( p – x )( l – x ) = Lrencana Jawab : ( p – x )( l – x ) ( 60 – x )(30 – x ) 1800 – 60x – 30x + x2 1800 – 90x + x2 1800 – 1000 – 90x + x2 800 – 90x + x2 800 – 80x – 10x + x2 80 ( 10 – x ) – x ( 10 – x ) ( 80 – x )(10 – x ) = Lrencana = 1000 = 1000 = 1000 =0 =0 =0 =0 =0 x = 80 atau x = 10 Diuji : Yang digunakan adalah x = 10 ( p – x )( l – x ) 60 – 10 . 30 – 10 50 . 20 2. Diket : ps ls t pa la = 50 cm = 40 cm =x =50 – 2x = 40 – 2x = Lrencana = 1000 = 1000 x ls x pp pa la ps
  • 3. Dit : Persamaan kuadrat…? Volume tempat air/balok…? Keterangan : ps = panjang seng Rumus : pa . la = La ls Jawab : pa . la = lebar seng t V =p.l.t = tinggi pa = panjang alas = La la = lebar alas ( 50 – 2x ) ( 40 – 2x ) = 200 2000 – 100x – 80x + 4x2 = 200 2000 – 180x + 4x2 = 200 2000 – 200 – 180x + 4x2 =0 1800 – 180x + 4x2 =0 450 – 45x + x2 =0 450 – 30x – 15x + x2 =0 30 ( 15 – x ) – x ( 15 – x ) =0 ( 30 – x ) ( 15 – x ) =0 La = luas alas x = 30 atau x = 15 Diuji : Yang digunakan adalah x = 15 pa = 50 – 2x la = 40 – 2 x = 50 – 2 . 15 = 50 – 30 = 40 – 30 = 20 cm t = 40 – 2 . 15 = 10 cm =x = 15 cm V =p .l .t = 20 . 10 . 15 = 3000 cm3
  • 4. 3. Diket : t = 3 cm Dit : jari-jari kerucut…? Rumus : V = V1 = V2 = Jawab t ( r + 242 ) 3 r2( 3 + 24 ) : V1 = ( r + 242 ) . 3 ( r + 242 ) . 3 3 ( r2+ 48r + 242 ) 3r2 + 3 . 48r + 3 . 24 .24 r2 + 48r + 24 . 24 0 0 0 0 r = 12 = V2 r2( 3 + 24 ) = r2 ( 3 + 24 ) = 27r2 = 27r2 = 9 r2 = 8 r2 - 48r - 24 . 24 = r2 - 6r - 24 . 3 = r2 - 6r - 72 = ( r – 12 ) ( r + 6 ) atau r = - 6 Jadi , jari – jari kerucut semula adalah 12 4. Diket : Dit : p1 = x jam P2 = x + 1 jam t1 + t2 = 1,2 jam t2 untuk mencetak 1 buku…? Keterangan : p1 : printer satu p2 : printer dua t1 : waktu printer satu t2 : waktu printer dua
  • 5. Jawab : = 1,2 = 1,2 = 1,2 = 1,2 x2 + x = 1,2 ( 2x + 1 ) x2 + x = 2,4x + 1,2 x2 + x – 2,4x – 1,2 =0 x2 – 1,4x – 1,2 =0 10x2 – 14x – 12x =0 10x2 – 20x + 6x – 12x = 0 10x ( x – 2 ) + 6 ( x – 2 ) = 0 ( 10x + 6 )( x – 2 ) =0 10x + 6 = 0 x–2=0 10x = -6 x x = -6 : 10 x = - 0,6 Yang digunakan adalah x = 2 p1 = x jam = 2 jam =2
  • 6. p2 = x jam + 1 jam = 2 jam + 1 jam = 3 jam Jadi , waktu yang diperlukan printer ke-dua untuk mencetak satu set buku adaah 3 jam.