Question

in \\( \triangle abc, \overline{bm} \\) is a median, \\( \overline{bh} \\) is an altitude, \\( bh = 6 \\), \\( m\angle bac = 30^{\circ} \\), and \\( bm=\frac{1}{2}ac \\). find area of \\( \triangle abc \\). answer: area of \\( \triangle abc = \\)

Explanation:

Step1: Determine the length of \(AB\)

In right - triangle \(ABH\), since \(\angle BAC = 30^{\circ}\) and \(BH\) is the altitude (\(BH = 6\)).
We know the property of a right - triangle: \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\sin\angle BAC=\frac{BH}{AB}\), and \(\sin30^{\circ}=\frac{1}{2}\).
So, \(\frac{BH}{AB}=\frac{1}{2}\), substituting \(BH = 6\), we get \(AB = 12\).

Step2: Use the property of the median

Since \(BM\) is a median (\(AM = MC\)) and \(BM=\frac{1}{2}AC\), then \(AM = BM=MC\).
In \(\triangle ABM\), because \(AM = BM\), \(\angle BAC=\angle ABM = 30^{\circ}\).
In \(\triangle ABC\), using the formula for the area of a triangle \(S=\frac{1}{2}AC\times BH\).
Since \(BM\) is a median and \(BM = AM\), \(\angle BMA=\angle BAC+\angle ABM=60^{\circ}\) (exterior angle property of a triangle). And since \(BM = AM\), \(\triangle ABM\) is an equilateral triangle. So \(AM = BM = AB = 12\). Then \(AC=2AM = 24\).

Step3: Calculate the area of \(\triangle ABC\)

The formula for the area of a triangle is \(S=\frac{1}{2}\times\text{base}\times\text{height}\). Here, the base \(AC\) and the height \(BH\).
Substitute \(AC = 24\) and \(BH = 6\) into the formula: \(S=\frac{1}{2}\times24\times6\).

$$S = 72$$

Answer:

\(72\)