this formula is for the area of an equilateral triangle.this is good when the height is not given.
1/2 bh is for all triangles.
2007-02-09 18:32:53
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answer #1
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answered by Blood 3
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It is the formula for the area of an equilateral triangle. But it is derived as a special case of the general formula for the area of a triangle, which is:
A = ½ bh.
In the case of an equilateral triangle, one of its sides, s, is the base of the triangle, and its height is [(√3)/2] s. We obtain this because when we drop a perpendicular from any vertex of an equilateral triangle, it bisects the angle from which it is dropped and also the side to which it is dropped. Thus it forms two right triangles whose hypotenuses equal s. And since the perpendicular bisects the vertex from which it is dropped, it forms two 30° angles there. Since the other two vertices of the equilateral triangle equal 60°, then we have two 30-60-90° triangles formed by dropping the perpendicular. We use the fact that in any 30-60-90° triangle the side opposite the 30° angle is always equal to half the length of the hypotenuse, in this case s. So the base leg of the triangle is s/2. Then we calculate the other leg, which is the height of the triangle, by using the Pythagorean Theorem. Thus h = √[(3 s²)/4] = [(√3)/2] s. When we substitute those values into the general formula, we get:
A = ½ (s) [(√3)/2] s = [(√3)/4] s².
2007-02-09 18:52:58
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answer #2
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answered by MathBioMajor 7
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The formula (root3/4)*side^2 is the formula for an equilateral triangle.
1/2 base*height is used for all other triangles.
Yet another formula is the Hero's formula:
root over [s(s-a)(s-b)(s-c)], where s is semi perimeter(perimeter/2),a,b and c are the sides.
2007-02-09 20:06:28
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answer #3
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answered by Raider 3
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That is the formula for the area of an equilateral triangle.
2007-02-09 18:33:33
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answer #4
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answered by bluefairy421 4
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Area for an equilateral triangle.
Here is the derivation.
Draw the equilateral triangle ABC.
Draw a perpendicular bisector of line AC from point B and label this as O.
Label AB = BC = AC = side (s)
Given: AO = CO = base (b) = (1/2)(s) = s/2
OB = height (h)
Using right triangle AOB (1/2 of the whole triangle):
OB^2 = s^2 - AO^2 (Pythagorean Theorem)
h^2 = s^2 - [(1/2)(s)]^2
h^2 = s^2 - (s^2)(1/4)
h = Sq. Rt[s^2 - s^2/4]
h = Sq. Rt.[(4s^2 - s^2)/4]
h = Sq. Rt.[(3s^2)/4]
Area (1) = (1/2)(b)(h)
Area (1) = (1/2)(AO){Sq. Rt (3s^2/4)}
Area (1) = (1/2)(s/2)[Sq. Rt. (3s^2)](1/2)
Area (1) = (1/2)(s/2)(Sq. Rt.3)(s)(1/2)
Using right triangle COB (1/2 of the whole triangle):
Area (2) = (1/2)(CO){Sq. Rt. (3s^2/4)}
Area (2) = (1/2)(s/2)(Sq. Rt. 3)(s)(1/2)
Area (Total) = [2] [Area (1)] = [2][Area (2)]
Area (Total) = [2][(1/2)(s/2)(Sq. Rt. 3)(s)(1/2)
Area (Total) = (s^2)(Sq. Rt. 3) / 4
or Simply:
Area = (s^2)(Sq. Rt. 3) / 4
Good luck.
Please email me if you have any question.
2007-02-09 19:54:01
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answer #5
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answered by Anonymous
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The specific and unique formula of equilateral triangle.
2007-02-09 18:38:43
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answer #6
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answered by honey 2
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Why is the formula Tn = (n/2) (n +1) works, show your proof
2015-09-27 16:26:01
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answer #7
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answered by Napetro 1
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hey.dis is the formula of equilateral triangle which has all sides equa and height sq.root3/2* side therefore formula=1/4sq.root3 *side sq.
2007-02-09 18:38:20
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answer #8
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answered by SS 2
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90 degree triangle... a2 + b2 = c2 where c2 is the hypotenuse (or the longest side) 45 degree triangle since both sides (L) are the same,the hypotenuse measures H=L√2
2016-03-29 00:35:52
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answer #9
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answered by Kathleen 4
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