Răspuns:
a) OD₁-raza tablei
DD₁-raza disc =5m
AD=2·DD₁=10m - latura patratului ABCD
OD -semidiagonala patratului ABCD, OD=AD·√2/2=5√2 m
OD₁=OD+DD₁=5√2+5 =5(1+√2) m
S - aria suprafetei de tabla
S = π·OD₁² = π·5²·(1+√2)².
b) S₁ -aria suprafetei inlaturate:
S₁ = S-4·( π·DD₁²) = π·5²·(1+√2)²- 4·π·5² = π·5²·[(1+√2)²-4] = π·5²·(1+2+2√2-4) = π·5²·(2√2-1)
c) 100·4·( π·DD₁²)/S = 100·( 4·π·5²)/[π·5²·(1+√2)²] = 400/(1+√2)² ≅ 400/(1+1,41)² = 400/5.8081 ≅ 68.86 %
