MON is a triangle such that MO = x+3, MN = 3 and ON = 6-x.
1) for what values of x do the lengths MO and ON exist? Justify
The sum of any two sides of a triangle must be
strictly greater than the third side:
MO+MN > ON MO+ON > MN MN+ON > MO
x+3 + 3 > 6-x x+3 + 6-x > 3 3+6-x > x+3
x+6 > 6-x 9 > 3 9-x > x+3
2x > 0 -2x > -6
x > 0 x < 3
So from those we get that 0 < x < 3
2) calculate x such that triangle MON is inscribed in a circle
of diameter the side [ON]If a triangle inscribed in a circle has a side which is a
diameter, then the triangle is a right angle.
We use the converse of the Pythagorean theorem, which states
that if the sides of a triangle a,b,c are such that a2+b2=c2,
then the triangle is a right triangle.
MO2 + MN2 = ON2
(x+3)2 + 32 = (6-x)2
x2+6x+9 + 9 = 36-12x+x2
x2+6x+18 = 36-12x+x2
18x = 18
x = 1
MO = x+3 = 1+3 = 4
MN = 3
ON = 6-x = 6-1 = 5
The triangle MON is a 3,4,5 right triangle.
Edwin
