tolong dijawab, klo bisa jwb dengan benar dan tepat saya jadiin yang terbaik(no 1 dan 3 yaaaa......)

1. a.
x^2 = 15^2 + 12^2
x^2 = 225 + 144
x^2 = 369
x = 369^(1/2)
x = 3 × (akar pangkat 2 dari 41)

b.
x^2 = 13^2 - 5^2
x^2 = 169 - 25
x^2 = 144
x = 144^(1/2)
x = 12

c.
a^2 = 10,6^2 - 5,6^2
a^2 = 112,36 - 31,36
a^2 = 81
a = 81^(1/2)
a = 9 inchi

d.
x^2 = 10,4^2 - 9,6^2
x^2 = 108,16 - 92,16
x^2 = 16
x = 16^(1/2)
x = 4m

e.
x^2 = 8^2 - 6^2
x^2 = 64 - 36
x^2 = 28
x = 28^(1/2)
x = 2 × (akar pangkat 2 dari 7)

f.
c^2 = 7,2^2 + 9,6^2
c^2 = 51,84 + 92,16
c^2 = 144
c = 144^(1/2)
c = 12 kaki

3.
a. x^2 = 20^2 - 12^2
x^2 = 400 - 144
x^2 = 56
x = 56^(1/2)
x = 2 × (akar pangkat 2 dari 14) cm

b.
x^2 = 35^2 + n^2
x^2 = 35^2 + (13^2 - 5^2)
x^2 = 1225 + (169 - 25)
x^2 = 1225 + 144
x^2 = 1369
x = 1369^(1/2)
x = 37 mm

Soal No. 1
a). a = 12, b = 15, c = x
[tex]c= \sqrt{a^2+b^2} \\ \\ c= \sqrt{12^2+15^2}= \sqrt{144 + 225}= \sqrt{369}= \sqrt{9 \times 41}= \sqrt{3^2} \times \sqrt{41} \\ \\ c=x=3 \sqrt{41}[/tex]

b) a = 5, b = x, c = 13
[tex]b= \sqrt{c^2-a^2} \\ \\ b= \sqrt{13^2-5^2}= \sqrt{169-25}= \sqrt{144}= \sqrt{12^2} \\ \\ b=x=12[/tex]

c) a = ?, b = 5,6 inch, c = 10,6 inch
[tex]a= \sqrt{c^2-b^2} \\ \\ a= \sqrt{(10,6)^2-(5,6)^2}= \sqrt{112,36-31,36}= \sqrt{81}= \sqrt{9^2} \\ \\ a=9\ inch[/tex]

d) a = ?, b = 9,6 m, c = 10,4 m
[tex]a= \sqrt{c^2-b^2} \\ \\ a= \sqrt{(10,4)^2-(9,6)^2}= \sqrt{108,16-92,16}= \sqrt{16}= \sqrt{4^2} \\ \\ a=4\ m[/tex]

e) a = 6, b = x, c = 8
[tex]b= \sqrt{c^2-a^2} \\ \\ b= \sqrt{8^2-6^2}= \sqrt{64-36}= \sqrt{28}= \sqrt{4 \times 7}= \sqrt{2^2} \times \sqrt{7} \\ \\ b=x=2 \sqrt{7} [/tex]

f) a = 9,6 feet, b = 7,2 feet, c = ?
[tex]c= \sqrt{a^2+b^2} \\ \\ c= \sqrt{(9,6)^2+(7,2)^2}= \sqrt{92,16 + 51,84} = \sqrt{144}= \sqrt{12^2} \\ \\ c=12\ feet[/tex]

Soal No. 3
a. a = x, b = 12 cm, c = 20 cm
[tex]a= \sqrt{c^2-b^2} \\ \\ a= \sqrt{20^2-12^2}= \sqrt{400-144}= \sqrt{256}= \sqrt{16^2} \\ \\ a=x=16\ cm[/tex]

b. 
- Segitiga Kecil : a = 5 mm , b = ? , c = 13 mm
[tex]b= \sqrt{c^2-a^2} \\ \\ b= \sqrt{13^2-5^2}= \sqrt{169-25}= \sqrt{144}= \sqrt{12^2}\\ \\ b=12\ mm[/tex]

- Segitiga Besar : a = 35 mm, b = 12 mm, c = x
[tex]c= \sqrt{a^2+b^2} \\ \\ c= \sqrt{35^2+12^2}= \sqrt{1.225 + 144} = \sqrt{1.369}= \sqrt{37^2} \\ \\ c=37\ mm[/tex]

***Semoga Terbantu***