Let \(x\) be the side of the triangle and \(a\) be the altitude.
Thus;
\(a = x - 3\)
\( x^2 = a^2 + (\frac{x}{2})^2\)
\( x^2 = (x-3)^2 + (\frac{x}{2})^2\)
\( x^2 = x^2 -6x + 9 + \frac{x^2}{4}\)
\( 0 = x^2 - 24x + 36\)
\( x = \frac{24 \pm \sqrt{24^2 -4(36)}}{2}\)
\( x = \frac{24 \pm \sqrt{24^2 -4(36)}}{2}\)
\(x_1 = 22.39 \; ft \)
\(x_2 = 1.61 \; ft\) ANSWER
Let \(x\) be the side of the triangle and \(a\) be the altitude.
Thus;
\(a = x - 3\)
We know that in an equilateral triangle the angles are \(60^o\) each.
\(\sin 60 = \frac{a}{x}\)
\(x \sin 60 = x-3\)
\(x [\sin 60 - 1] = -3\)
\( x = \frac{3}{1-\sin 60}\)
\(x = \dfrac{3}{\frac{2\pm\sqrt{3}}{2}}\)
\(x_1 = 22.39 \; ft \)
\(x_2 = 1.61 \; ft\) ANSWER
Let \(x\) be the side of the triangle and \(a\) be the altitude.
Thus;
\(a = x - 3\)
We know that the area of a triangle is \(A_T = 0.5 x(a)\) and if it's an equilateral triangle it would be \(A_T = 0.5 x^2 \sin \theta\). Also, for an equilateral triangle the angles are \(60^o\)
Thus; \(A_T = 0.5x(a) = 0.5 x^2 \sin 60 \)
\((x-3) = x \sin 60\)
\(x-3 = x \frac{\sqrt{3}}{2}\)
\(x = \dfrac{3}{\frac{2\pm \sqrt{3}}{2}}\)
\(x_1 = 22.39 \; ft \)
\(x_2 = 1.61 \; ft\) ANSWER
