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Topic:
Can I get some help please.
Replies:
4
Last Post:
Oct 6, 2009 4:09 PM




Re: Can I get some help please.
Posted:
Oct 6, 2009 3:38 PM


Hi Tony,
If all the sides have length 72, then there are a couple of ways you can get the area. One is to use Heron's formula. In that case, the semiperimeter s 3*72/2 = 108. And the area is
sqrt(s(sa)(sb)(sc))
= sqrt(108(10872)(10872)(10872))
= sqrt(108*36^3)
= sqrt(3*36^4)
= 36^2 * sqrt(3)
= 2245 (approx)
You can also drop an altitude from any of the vertices. The length of the altitude will be
72 * sin(60)
which is the height of the triangle; so the area will be
1/2 * base * height
= 1/2 * 72 * 72 * sin(60)
= 1/2 * 72 * 72 * sqrt(3)/2
= 36^2 * sqrt(3)
same as before.
I don't understand where you're getting the numbers 37 and 73, or why you think multplying them would give you the area. Can you say more about your reasoning there?
Looking ahead, what you can do is divide the triangle into three smaller triangles, each with a vertex at the center of the triangle. Each of them will be a 3030120 triangle. Divide one in two, and you have a 306090 triangle. The hypotenuse of that triangle will be the radius of the circle you're looking for.



