The document summarizes three word problems involving quadratic equations and their solutions: 1) A problem about the area of a rectangle with length p=60m and width l=30m having area L=1000m^2. The solutions are x=80m or x=10m. 2) A problem about the area of a trapezoid with bases of 50-2x cm and 40-2x cm and height x cm having area of 200cm^2. The solutions are x=30cm or x=15cm. 3) A problem about the volume of a cylinder with radius r and height t=3cm having volume V=27cm^3. The solutions are

1. NAMA KELOMPOK :
NO ABSEN :
NI WAYAN NANIK ASTITI GIRI
(08)
NI PUTU DESI ARYA DEWI
(12)
DINDA SIONA EKSUBERANTI
(20)
NI LUH PUTU TIA PUSPITA DEWI
(32)
MATERI PEMBELAJARAN : MENENTUKAN KONSEP PERSAMAAN KUADRAT
SMK NEGERI 1 TABANAN
TAHUN AJARAN 2013/2014

2. 1. Diket : p = 60 m
l = 30 m
= 1000 m
Rumus : ( p – x )( l – x ) = L
Jawab :
( p – x )( l – x )
=L
( 60 – x )(30 – x )
= 1000
1800 – 60x – 30x +
= 1000
1800 – 90x +
= 1000
1800 – 1000 – 90x +
=0
800 – 90x +
=0
800 – 80x – 10x +
=0
80 ( 10 – x ) – x ( 10 – x ) = 0
( 80 – x )(10 – x )
=0
x = 80 atau x = 10
p
l
( p – x )( l – x )
=
60 – 10 x 30 – 10 = 1000
50 x 20
= 1000
2. Diket :
= 50 cm
= 40 cm
t =x
= 50 – 2x
= 40 – 2x
Keterangan :
= panjang seng
= lebar seng
t = tinggi
x
( 50 – 2x ) ( 40 – 2x )
=
= panjang alas
= 200
= lebar alas
2000 – 100x – 80x + 4
= 200
2000 – 180x + 4
= 200
2000 – 200 – 180x + 4
=0
1800 – 180x + 4
=0
450 – 45x +
=0
= luas alas

3. 450 – 30x – 15x +
=0
30 ( 15 – x ) – x ( 15 – x 0 = 0
( 30 – x ) ( 15 – x )
x
=0
x
x
X = 30 atau x = 15
x
x
x
x
= 50 – 2x
= 40 – 2 . 15
= 50 – 30
= 40 – 30
= 20 cm
t
= 40 – 2 x
= 50 – 2 . 15
x
= 10 cm
=x
V=p.l.t
= 15 cm
= 20 . 10 . 15
= 3000 c
3. Diket : t = 3 cm
V =
=
t
(
=
.3
(3+4 )
=
.3
=
t
( 3 + 24 )
.3
=
( 3 + 24 )
3 ( + 48r +
) = 27
3 + 3 . 48r + 3 . 24 .24 = 27
+ 48r + 24 . 24 = 9
8 – 48r + 24 . 24
0 =
- 6r – 24 . 3
0=
- 6r - 72
0 = ( r – 12 ) ( r + 6 )
r = 12 atau r = - 6
r