1.
5² + (2√6)² = x ²
25 + 4 * 6 = x²
25 + 24 = x²
49 = x²
x = √49
x = 7
zad 2.
8² + x² = 12 ²
64 + x² = 144
x² = 80
x = √80 = 2√20
obwód = 2 * 8 + 2* 2√20 = 16 + 4√20
Ob= 16 + 4√20 [ cm]
Pole:
8 * 2√20 = 16√20 [cm²]
Zadanie 1
z twierdzoenia pitagorasa
x- przeciwprostokątna
x²=5²+(2√6)²
x²=25+2*6
x²=25+12
x²=37
x=√37
Zadanie 2
a=8cm
b= ?
12²=8²+b²
144=64+b²
b²=80
b=√80
P=8√80
Ob=16 + 2√80
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1.
5² + (2√6)² = x ²
25 + 4 * 6 = x²
25 + 24 = x²
49 = x²
x = √49
x = 7
zad 2.
8² + x² = 12 ²
64 + x² = 144
x² = 80
x = √80 = 2√20
obwód = 2 * 8 + 2* 2√20 = 16 + 4√20
Ob= 16 + 4√20 [ cm]
Pole:
8 * 2√20 = 16√20 [cm²]
Zadanie 1
z twierdzoenia pitagorasa
x- przeciwprostokątna
x²=5²+(2√6)²
x²=25+2*6
x²=25+12
x²=37
x=√37
Zadanie 2
a=8cm
b= ?
12²=8²+b²
144=64+b²
b²=80
b=√80
P=8√80
Ob=16 + 2√80