W trójkąt równoboczny wpisano koło o promieniu r=8.O ile procent pole trójkąta jest większe od pola koła.
Daje naj
r = 8
r = (1/3)*h
h = 3*r = 3*8 = 24
================
oraz
h = a p(3)/2
2 h = a p(3)
a = (2 h)/p(3)
a = (2*24)/ p(3) = 48/ p(3) = 16 p(3)
=================================
Pole trójkąta
Pt = (1/2) a*h = (1/2) *16 p(3)*24 = 192 p(3)
pole koła
Pk = pi r^2 = pi * 8^2 = 64 pi
============================
Pt - Pk = 192 p(3) - 64 pi = 332,55 - 200,96 = 131,59
131,59/200,96 = 0,6548 = 65,5 %
Odp. 65,5 %
===============
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r = 8
r = (1/3)*h
h = 3*r = 3*8 = 24
================
oraz
h = a p(3)/2
2 h = a p(3)
a = (2 h)/p(3)
a = (2*24)/ p(3) = 48/ p(3) = 16 p(3)
=================================
Pole trójkąta
Pt = (1/2) a*h = (1/2) *16 p(3)*24 = 192 p(3)
oraz
pole koła
Pk = pi r^2 = pi * 8^2 = 64 pi
============================
Pt - Pk = 192 p(3) - 64 pi = 332,55 - 200,96 = 131,59
131,59/200,96 = 0,6548 = 65,5 %
Odp. 65,5 %
===============