Question

1. what is the area of triangle abc? figure 1 (triangle with angles at a and c as 60°, ab=20, bd perpendicular to ac at d) options: 200√3, 200, 100√3, 100

Explanation:

Step1: Analyze triangle ABD

In right triangle \(ABD\), \(\angle A = 60^\circ\), \(AB = 20\). We can find the height \(BD\) and base \(AD\) using trigonometric ratios. \(\sin60^\circ=\frac{BD}{AB}\), so \(BD = AB\times\sin60^\circ = 20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\). \(\cos60^\circ=\frac{AD}{AB}\), so \(AD = AB\times\cos60^\circ = 20\times\frac{1}{2} = 10\).

Step2: Analyze triangle ABC

Since \(\angle A=\angle C = 60^\circ\), triangle \(ABC\) is isosceles, so \(AC = 2\times AD=20\). The area of a triangle is \(\frac{1}{2}\times base\times height\). Here, base \(AC = 20\), height \(BD = 10\sqrt{3}\). So area \(=\frac{1}{2}\times20\times10\sqrt{3}=100\sqrt{3}\). Wait, no, wait. Wait, in triangle \(ABC\), since \(\angle A=\angle C = 60^\circ\), then \(\angle B = 60^\circ\), so it's an equilateral triangle? Wait, \(AB = 20\), if it's equilateral, then all sides are 20. Wait, maybe my first approach was wrong. Wait, let's re - check.

Wait, in triangle \(ABC\), \(\angle A = 60^\circ\), \(\angle C=60^\circ\), so \(\angle B = 60^\circ\), so triangle \(ABC\) is equilateral with side length 20? Wait, no, the length of \(AB\) is 20. Wait, maybe the height is calculated as \(h = AB\times\sin60^\circ\) when we consider the base \(AC\). Wait, no, let's use the formula for the area of a triangle with two sides and included angle. The formula is \(\frac{1}{2}ab\sin C\). In triangle \(ABC\), \(AB = BC = 20\) (since \(\angle A=\angle C\)), and \(\angle B = 60^\circ\). So area \(=\frac{1}{2}\times AB\times BC\times\sin B=\frac{1}{2}\times20\times20\times\sin60^\circ=\frac{1}{2}\times400\times\frac{\sqrt{3}}{2}=100\sqrt{3}\). Wait, but earlier when I thought \(AC = 20\), that was a mistake. Wait, no, in the right triangle \(ABD\), \(AD = 10\), so \(AC = 2\times AD = 20\), so base \(AC = 20\), height \(BD = 10\sqrt{3}\), so area \(\frac{1}{2}\times20\times10\sqrt{3}=100\sqrt{3}\).

Wait, but let's check again. Wait, the triangle \(ABC\): angles at \(A\) and \(C\) are \(60^\circ\), so angle at \(B\) is \(60^\circ\), so it's equilateral. So side length is 20. The area of an equilateral triangle is \(\frac{\sqrt{3}}{4}a^{2}\), where \(a = 20\). So area \(=\frac{\sqrt{3}}{4}\times20^{2}=\frac{\sqrt{3}}{4}\times400 = 100\sqrt{3}\). Yes, that matches.

Answer:

\(100\sqrt{3}\) (corresponding to the option \(100\sqrt{3}\))