SOLUTION: 72. Geometry - Write the area A of an equilateral triangle as a function of the length s of its sides.
I need help. Thanks
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Question 268519: 72. Geometry - Write the area A of an equilateral triangle as a function of the length s of its sides.
I need help. Thanks
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Write the area A of an equilateral triangle as a function of the length s of its sides.
I need help. Thanks
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Draw the triangle; label each side "2s".
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Draw the altitute; it will bisect the base into s and s.
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Use Pythagoras to determine the length of the altitude:
h^2 = s^2 + (2s)^2
h^2 = s^2 + 4s^2
h^2 = 5s^2
h = s*sqrt(5)
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Area of the triangle = (1/2)(height)(base)
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A = (1/2)(s*sqrt(5)(2s)
A = (1/2)(sqrt(5)((1/2)side)(side)
A = (sqrt(5)/4)(side)^2
==============
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The bisector of an interior angle of an equilateral triangle and the perpendicular bisector of the side opposite the bisected angle are contained in the same line, and the segment from the vertex of the bisected angle to the midpoint of the opposite side, being perpendicular to that side, is an altitude of the triangle. Therefore, the area is equal to the measure of the bisected side times the measure of the altitude, quantity divided by 2. Now all we need is the measure of the altitude in terms of the measure of a side.
Since the altitude is perpendicular to the bisected side and passes through the opposite vertex, the altitude forms one leg of a right triangle where the other leg is one-half the measure of a side (the altitude is a bisector of that side) and the hypotenuse is one of the sides. Pythagoras says:
\ =\ \sqrt{s^2\ -\ \left(\frac{s}{2}\right)^2}\ =\ \sqrt{s^2\ -\ \frac{s^2}{4}}\ =\ \sqrt{\frac{3s^2}{4}}\ =\ \frac{s\sqrt{3}}{2})
Then the area is:
\ =\ \frac{sH(s)}{2}\ =\ \left(\frac{s}{2}\right)\left(\frac{s\sqrt{3}}{2}\right)\ =\ \frac{s^2\sqrt{3}}{4})
John
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