Dane: a = 5 , r = 2
CD = r AD = AE = a − r CF = r BF = BE = b − r
AB = AE + EB = a − r + b − r = a + b − 2*r = c
Z tw. Pitagorasa
a2 + b2 = c2
a2 + b2 = (a + b − 2*r)2 podstawiam dane
52 + b2 = (5 + b − 2*2)2
25 + b2 = (b + 1)2
25 + b2 = b2 + 2*b + 1 ⇒ 2*b = 24 ⇒ b = 12
Szukane pole
| 1 | 1 | |||
P = | *a*b = | *5*12 = 30 | ||
| 2 | 2 |