Guru SMPN 2 Jember: http://olimattohir.blogspot.co.id/

1. http://m2suidhat.blogspot.com/
http://m2suidhat.blogspot.com/
Mathematics Sport
𝟏𝟓
𝟏
𝟐
+ 𝟐
𝟐
𝟑
− 𝟗
𝟏
𝟒

𝟑𝟏
𝟐
+
𝟖
𝟑
−
𝟑𝟕
𝟒
=
𝟏𝟖𝟔
𝟏𝟐
+
𝟑𝟐
𝟏𝟐
−
𝟏𝟏𝟏
𝟏𝟐
=
𝟐𝟏𝟖
𝟏𝟐
−
𝟏𝟏𝟏
𝟏𝟐
=
𝟏𝟎𝟕
𝟏𝟐
= 8
𝟏𝟏
𝟏𝟐
36  8
x  24
𝒙
𝟑𝟔
=
𝟖
𝟐𝟒
 x = 𝟑𝟔 ×
𝟖
𝟐𝟒
 x =
𝟑𝟔
𝟑
 x = 12 buah
𝟏𝟐𝟓
𝟐
𝟑 = 𝟓 𝟑
𝟐
𝟑 = 𝟓 𝟐
= 𝟐𝟓
𝟔𝟎 : 𝟓  𝟏𝟐 × 𝟓 : 𝟓
= 𝟏𝟐 × 𝟓 : 𝟓
= 𝟏𝟐
= 𝟒 × 𝟑
= 2 𝟑
𝟔𝟎 : 𝟓  𝟔𝟎: 𝟓
= 𝟏𝟐
= 𝟒 × 𝟑
= 2 𝟑
atau
Paket 1

2. http://m2suidhat.blogspot.com/
𝟐
𝟑

𝟐
𝟑
×
𝟑
𝟑
=
𝟐 𝟑
𝟑
(i) Besar bunga = 920.000 – 800.000 = 120.000
(ii) Besar bunga =
𝒃𝒖𝒍𝒂𝒏
𝟏𝟐
× 9% × 800.000
Sehingga: 120.000 =
𝒃𝒖𝒍𝒂𝒏
𝟏𝟐
×
𝟗
𝟏𝟎𝟎
× 800.000  120 =
𝒃𝒖𝒍𝒂𝒏
𝟒
× 3 × 8
 bulan =
𝟏𝟐𝟎
𝟑×𝟐
= 20 bulan
b =
𝑼 𝟕−𝑼 𝟐
𝟕−𝟐
=
𝟑𝟏−𝟔
𝟓
=
𝟐𝟓
𝟓
= 5
U2 = a + (2-1)5
6 = a + 5
a = 1
U40 = a + (40 – 1)5
U40 = 1 + (39)5
U40 = 1 + 195
U40 =196
b =
𝑼 𝟕−𝑼 𝟑
𝟕−𝟑
=
𝟑𝟖−𝟏𝟖
𝟒
=
𝟐𝟎
𝟒
= 5
U3 = a + (3-1)5
18 = a + 10
a = 8
S24 =
𝟐𝟒
𝟐
(2×8 + (24 – 1)5)
S24 = 12(16 + (23)5)
S24 = 12(16 + 115)
S24 = 12(131)
S24 =1572
Barisan gaji: 3.000.000, 3.500.000, 4.000.000, 4.500.000,....
S10 =
𝟏𝟎
𝟐
(2×3.000.000 + (9)500.000)
S10 = 5(6.000.000 + 4.500.000)
S10 = 5(10.500.000)
S10 = 52.500.000
(i) 15x2
y – 20xy2
= 5xy(3x – 4y)
(ii) p2
– 16 = (p + 4)(p – 4)
(iii) 3a2
+ 8a – 3 = (3a – 1)(a + 3)
Paket 1

3. http://m2suidhat.blogspot.com/
3x + 6 = 5x + 20
6 – 20 = 5x – 3x
– 14 = 2x
– 7 = x
x = – 7
K = 2(p + l)
44 = 2(p + l)
22 = (p + l)
22 = (3x + 4) + (2x + 3)
22 = 5x + 7
22 – 7 = 5x
15 = 5x
3 = x
x = 3
p = 3x + 4 = 3(3) + 4 = 9 + 4 = 13
l = 2x + 3 = 2(3) + 3 = 6 + 3 = 9
Banyak anggota D = 6
Banyak himpunan bagian dari D adalah 26
= 64
P C40
12
23-12=11
40 – (11 + 12) = 40 – 23 = 17
Menulis cerpen = 17 + 12 = 29
f(x) = 8 – 2x  f(k) = 8 – 2k
 –10 = 8 – 2k
 –18 = –2k
 9 = k
 k = 9
Jadi, x + 12 = – 7 + 12 = 5
Paket 1

4. http://m2suidhat.blogspot.com/
m1 =
𝟐−𝟕
𝟎+𝟏
= – 5
m1×m2 = –1
–5 × m2 = –1
m2 =
𝟏
𝟓
m = 2
Memotong sumbu-y di titik (0, 2)
m = −
𝟑
𝟐
Memotong sumbu-y bukan di titik (0, 2)
m =
𝟐
𝟏
= 2
Memotong sumbu-y bukan di titik (0, 2)
m =
𝟏
𝟏
= 1
Memotong sumbu-y bukan di titik (0, 2)
m =
𝟐
𝟏
= 2
Memotong sumbu-y di titik (0, 2)
A. m2 = 5
B. m2 =
𝟏
𝟓
C. m2 = – 5
D. m2 = – 5
m1 =
−𝟏𝟑−𝟏
−𝟒−𝟑
=
−𝟏𝟒
−𝟕
= 2
y – y1 = m(x – x1), (ambil x1 = 3 dan y1 = 1)
y – 1 = 2(x – 3)
y – 1 = 2x – 6
y = 2x – 5
y = 2x – 5 , titik A (10, p)
p = 2(10) – 5
p = 20 – 5
p = 15
x – 3y – 5 = 0  ×2  2x – 6y – 10 = 0
2x – 5y – 9 = 0  ×1  2x – 5y – 9 = 0
–y – 1 = 0
y = –1
sehingga x – 3y – 5 = 0
x – 3(–1) – 5 = 0
x + 3 – 5 = 0
x – 2 = 0
x = 2
Nilai 3x + 2y = 3(2) + 2(–1)
= 6 – 2
= 4
Paket 1

5. http://m2suidhat.blogspot.com/
5a + 3j = 79.000  ×2  10a + 6j = 158.000
3a + 2j = 49.000  ×3  9a + 6j = 147.000
a = 11.000
Panjang tali = 𝟏𝟓𝟎 𝟐 + 𝟏𝟓𝟎 𝟐 = 𝟐 × 𝟏𝟓𝟎 𝟐 =𝟏𝟓𝟎 𝟐
Panjang perkiraan dengan 𝟐 𝟏, 𝟒𝟏 adalah 150 × 1,41 = 211,5  212
K L
N M
B
C
D
A
K L
N M
B
CD
A cm8
cm8
cm10
cm10
cm10
Luas daerah yang diarsir =
𝟏
𝟒
× Luas persegi KLMN
=
𝟏
𝟒
× (8)2
= 16 cm2
Paket 1

6. http://m2suidhat.blogspot. com/
(4 + 4 + 6 + 6) × 2 = (20) × 2 = 40 cm
4 cm
4 cm
6 cm
6 cm
atau
1
2
4
3
(i) 1 dengan 3
(ii) 2 dengan 4
(iii) 12 dengan 34
(iv) 14 dengan 23
ada 4 pasang
x

Q R
P
S
T
R
10 cm
15 cm
6 cm
𝑻𝑹
𝑸𝑹
=
𝑹𝑺
𝑷𝑹

𝑻𝑹
𝟏𝟎
=
𝟔
𝟏𝟓
 TR = 𝟏𝟎 ×
𝟔
𝟏𝟓
 TR = 4 cm
Jadi, PT = PR – TR
= 15 – 4
PT = 11 cm
Paket 1

7. http://m2suidhat.blogspot.com/
D A
E
B C
E
x  x 
𝑨𝑫
𝑩𝑪
=
𝑨𝑬
𝑬𝑪
=
𝑫𝑬
𝑩𝑬
A + B = 1800
(2x + 30) + (5x + 10) = 180
7x + 40 = 180
7x = 140
x = 20
K L
M
4
1 2
3
Jadi, B = (5x + 10) = 5(20) + 10
= 100 + 10
B = (5x + 10) = 1100
K L
M
3
(i)
K L
M
2
(ii)
K L
M
1
(iii)
K L
M
4
(iv)
Panjang busur AB =
𝟕𝟐 𝐨
𝟑𝟔𝟎 𝐨
× 2 r =
𝟏
𝟓
× 2 ×
𝟐𝟐
𝟕
× 14
=
𝟏
𝟓
× 44 × 2
=
𝟖𝟖
𝟓
= 17, 6 cm
Paket 1

8. http://m2suidhat.blogspot.com/
d2
= p2
– (r1 + r2)2
 p2
= d2
+ (r1 + r2)2
= 162
+ (8 + 4)2
= 256 + (12)2
= 256 + 144
= 400
 p = 20
Ada 18 (6×3) rusuk dan 8 (6 + 2) sisi
V = La × t
=
𝟏𝟖+𝟏𝟐
𝟐
× 𝟏𝟎 × 20
= (30 × 5) × 20
= 150 × 20
V = 3000
Dengan menggunakan teorema pythagoras
didapat panajang TE = 10 cm
L = luas alas + 4 × luas sisi tegak
= 122
+ 4 ×
𝟏
𝟐
× 12 × 10
= 144 + 240
L = 384 cm212 cm
8 cm
EO
Paket 1

9. http://m2suidhat.blogspot.com/
LP = Luas tabung tanpa tutup + luas setengah bola
= (πr2
+ 2πrt) + 2πr2
= (
𝟐𝟐
𝟕
×72
+ 2×
𝟐𝟐
𝟕
×7×20) + 2×
𝟐𝟐
𝟕
×72
= (22×7 + 2×22×20) + 2×22×7
LP = 1.342
4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10
Median =
𝟔+𝟕
𝟐
= 6,5
Bisa jadi, rata-ratanya = 130
Bisa jadi, rata-ratanya = 130
Bisa jadi, rata-ratanya = 130
Bisa dipastikan, rata-ratanya = 130
Rata-rata =
𝟐𝟑×𝟏𝟑𝟎+𝟏𝟑𝟑+𝟏𝟐𝟕
𝟐𝟑+𝟏+𝟏
=
𝟑𝟐𝟓𝟎
𝟐𝟓
= 130
180 130 525 640 360 200  2035
 30
Rata-rata =
𝟐𝟎𝟑𝟓
𝟑𝟎
= 67, 83
Banyak siswa yang nilainya lebih dari rata-rata = 30 – 9 = 21
Paket 1

10. http://m2suidhat.blogspot.com/
Senin – Selasa dan Kamis – Jum’at  penurunan 1000
Selasa – Rabu  kenaikan 2000
Rabu – Kamis dan Jum’at – Sabtu  uang saku tetap
Senin – Selasa  penurunan 1000
Nomor bola kurang dari 5  1, 2, 3, 4  ada 4
P =
𝒏𝒐𝒎𝒐𝒓 𝒃𝒐𝒍𝒂 𝒌𝒖𝒓𝒂𝒏𝒈 𝒅𝒂𝒓𝒊 𝟓
𝒃𝒂𝒏𝒚𝒂𝒌 𝒏𝒐𝒎𝒐𝒓 𝒃𝒐𝒍𝒂
=
𝟒
𝟗
Disusun oleh : Mohammad Tohir
Jika ada saran, kritik maupun masukan
silahkan kirim ke- My email: mohammadtohir@yahoo.com
Terima kasih.
My blog : http://m2suidhat.blogspot.com/
http://matematohir.wordpress.com/
Paket 1