What radius of a circle is required to inscribe an equilateral triangle with an area of 270.633 cm2 and an altitude of 21.65 cm? (round to nearest tenth) a) 10.8 cm b) 14.4 cm c) 15.4 cm d) 16.4 cm can you also explain how you got that answer?

We know that
 the distance from the centroid of the triangle to one of the vertices is the radius of the circle required to inscribe an equilateral triangle.

[distance centroid of the triangle to one of the vertices]=(2/3)*h
h=the altitude  of the equilateral triangle> 21.65 cm
so
[distance centroid of the triangle to one of the vertices]=(2/3)*21.65
[distance centroid of the triangle to one of the vertices]=14.43 cm

the radius is equal to the distance of the centroid of the triangle to one of the vertices
hence
the radius is 14.4 cm

the answer is
the radius is 14.4 cm

The point (5, 10) will work for the rule y=x+5.y= 10 x= 510=5+510=10therefore the ordered pair (5, 10) works for the rule