Question

Zad w załącz. Bardzo proste. Całe(C)

Answer 1

Z twierdzenia pitagorasa obliczamy boki tych kwadratów:
a^2 + b^2 = c^2
I.
12^2 + 5^2 = c^2
144 + 25 = c^2
c^2 = 169
c = √ 169
c = 13

P = a^2
P = 13^2
P = 169

II.
5^2 + 10^2 = c^2
25 + 100 = c^2
c^2 = 125
c = √ 125
c = √ 5 * 25
c = 5 √ 5

P = c^@
P = (5 √ 5)^2
P = 25 * 5
P = 125

III.
(3 √ 2)^2 + (3 √ 14)^2 = c^2
9 * 2 +  9 * 14 = c^2
18 + 126 = c^2
c^2 = 144
c = √ 144
c = 12

P = 12^2
P = 144

Answer 2

12^2+5^2=a^2 (a-nok kwadratu)
144+25=a^2
169=a^2
a=13 (a>0)
P=13^2=169

5^2+10^2=b^2(b-bok kwadratu ,b>0)
25+100=b^2
125=b^2
b=pierw z 125
P=(pierw z 125)^2=125

(3pierw z 2)^2+(3pierw z 14)^2=c^2(c-bok kwadratu ,c>0)
18+126=c^2
144=c^2
c=12
P=12^2=144