Answer is C. Sum of distances of any vertex of a triangle of side b to sides of triangle side a, is a constant and equals to altitude h = sqrt(3)*a/2, then: x+y+z+sqrt(3)*b/2=sqrt(3)*a/2 -> x+y+z=sqrt(3)*(a-b)/2 = 3/2 = 1.5 If you notice that x+y+z is a constant for two triangles with sides a and b, you can move inside triangle of side b to any vertex of exterior triangle side a, to obtain for example y=z=0 and x=(a-b)*sin(60º)=3/2.
1.5:)
ReplyDeleteAnswer is C.
ReplyDeleteSum of distances of any vertex of a triangle of side b to sides of triangle side a, is a constant and equals to altitude h = sqrt(3)*a/2, then:
x+y+z+sqrt(3)*b/2=sqrt(3)*a/2 -> x+y+z=sqrt(3)*(a-b)/2 = 3/2 = 1.5
If you notice that x+y+z is a constant for two triangles with sides a and b, you can move inside triangle of side b to any vertex of exterior triangle side a, to obtain for example y=z=0 and x=(a-b)*sin(60º)=3/2.
MIGUE.
Root3 sin60 = 1.5
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