Sagot :
Given:
side = 12
see figure:
a right triangle is formed.
where the hypotenuse is. h - r
and the other sides are r and 12/2
[tex]h= \sqrt{6^2+12^2} [/tex]
h=10.39
to get the radius we use the Pythagorean theorem.
[tex]10.39-r= \sqrt{(r)^2+(6)^2} [/tex]
[tex]r=3.46[/tex]
side = 12
see figure:
a right triangle is formed.
where the hypotenuse is. h - r
and the other sides are r and 12/2
[tex]h= \sqrt{6^2+12^2} [/tex]
h=10.39
to get the radius we use the Pythagorean theorem.
[tex]10.39-r= \sqrt{(r)^2+(6)^2} [/tex]
[tex]r=3.46[/tex]
r = 2 √3
See attached for solution. :)
Btw, I used trig functions, because it is the easiest way to solve it. Haha. I hope you're familiar with that.
<XBO is 1/2 of <XBA. That is why it's 30º
See attached for solution. :)
Btw, I used trig functions, because it is the easiest way to solve it. Haha. I hope you're familiar with that.
<XBO is 1/2 of <XBA. That is why it's 30º