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: Use property of right - triangle

Since \(BH\) is an altitude (\(\angle BHA = 90^{\circ}\)) and \(\angle BAC=30^{\circ}\), in right - triangle \(ABH\), we know that \(\sin\angle BAC=\frac{BH}{AB}\). Given \(BH = 6\) and \(\angle BAC = 30^{\circ}\), and \(\sin30^{\circ}=\frac{1}{2}\). So, \(\sin30^{\circ}=\frac{BH}{AB}\), then \(AB = 12\).

Step2: Use property of median and given relation

Since \(BM=\frac{1}{2}AC\) and \(M\) is the mid - point of \(AC\) (because \(BM\) is a median, so \(AM = MC\)), let \(AM=x\), then \(AC = 2x\) and \(BM=x\). In \(\triangle ABM\), \(BM = AM\), so \(\angle ABM=\angle BAC = 30^{\circ}\). Then \(\angle BMC=\angle ABM+\angle BAC=60^{\circ}\) (exterior angle of \(\triangle ABM\)).

Step3: Find length of \(AC\)

In right - triangle \(BHC\), in \(\triangle BMC\), since \(BM = MC\) (because \(BM=\frac{1}{2}AC\) and \(AM = MC\)) and \(\angle BMC = 60^{\circ}\), \(\triangle BMC\) is an equilateral triangle. So \(AC=AB = 12\).

Step4: Calculate the area of \(\triangle ABC\)

The formula for the area of a triangle is \(S=\frac{1}{2}\times base\times height\). Here, the base \(AC\) and the height \(BH\). Substitute \(AC = 12\) and \(BH = 6\) into the formula \(S=\frac{1}{2}\times AC\times BH\). So \(S=\frac{1}{2}\times12\times6\).

Answer:

\(36\)