Question

isosceles △abc (ac = bc) has base angles of 30° and cd is the median to base. how long are the legs of △abc, if the sum of the perimeters of △acd and △bcd is 20 cm more than the perimeter of △abc? answer: the legs are cm and cm

Explanation:

Step1: Analyze the perimeters of triangles

Let \(AC = BC = x\) and \(AB=y\). The perimeter of \(\triangle ABC\) is \(P_{ABC}=2x + y\). Since \(CD\) is the median to the base \(AB\) (so \(AD=BD=\frac{y}{2}\)), the perimeter of \(\triangle ACD\) is \(P_{ACD}=x + \frac{y}{2}+CD\) and the perimeter of \(\triangle BCD\) is \(P_{BCD}=x+\frac{y}{2}+CD\). The sum of the perimeters of \(\triangle ACD\) and \(\triangle BCD\) is \(P_{ACD}+P_{BCD}=2x + y+ 2CD\).
We know that \(P_{ACD}+P_{BCD}=P_{ABC}+20\). Substituting the expressions for the perimeters, we get \(2x + y+2CD=(2x + y)+20\), so \(CD = 10\).

Step2: Use the property of isosceles - triangle and angles

In isosceles \(\triangle ABC\) with base - angles of \(30^{\circ}\), and \(CD\perp AB\) (a property of the median in an isosceles triangle). In right - triangle \(ACD\), \(\cos30^{\circ}=\frac{CD}{AC}\).
Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\) and \(CD = 10\), and \(\cos30^{\circ}=\frac{CD}{AC}\), we have \(AC=\frac{CD}{\cos30^{\circ}}\).

Step3: Calculate the length of the legs

Substitute \(CD = 10\) into the formula \(AC=\frac{CD}{\cos30^{\circ}}\). Then \(AC=\frac{10}{\frac{\sqrt{3}}{2}}=\frac{20\sqrt{3}}{3}\). Since \(AC = BC\), both legs have a length of \(\frac{20\sqrt{3}}{3}\) cm.

Answer:

\(\frac{20\sqrt{3}}{3}\), \(\frac{20\sqrt{3}}{3}\)