r+h = 10 r = 10−h
l = 2√13
h2 + r2 = l2
h2 + (10−h)2 = 52
h2 + 100 − 20h − 52 = 0
h2 −20h + 48 = 0
Δ = 208
√Δ = 4√13
h1 = 10−2√13
h2 = 10+2√13
r1 = 10 −(10−2√13) = 2√13
r2 = 10−(10+2√13) = −2√13 → odrzucamy
Pc = πr( r+l) =
| πr2*h | ||
V = | = | |
| 3 |