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Problem 27

# $$A B C$$ is an equilateral triangle such that the vertices $$B$$ and $$C$$ lie on two parallel lines at a distance 6. If A lies between the parallel lines at a distance 4 from one of them, then the length of a side of the equilateral triangle is : (a) 8 (b) $$\sqrt{\frac{88}{3}}$$ (c) $$\frac{4 \sqrt{7}}{\sqrt{3}}$$ (d) none of these

Expert verified
$$\sqrt{\frac{88}{3}}$$
See the step by step solution

## Step 1: Understanding the Problem

The vertices $$B$$ and $$C$$ of an equilateral triangle $$ABC$$ lie on two different parallel lines, with $$A$$ being between them. The distance between these parallel lines is given as 6 units, while the distance of $$A$$ from one of the lines is given as 4 units. Considering these distances as vertical heights, we can understand that the distance of $$A$$ from the other line is $$6-4=2$$ units.

## Step 2: Drawing the Triangle and Identifying the Altitude

Draw the triangle such that $$A$$ is the top vertex and the base $$BC$$ lies along one of the parallel lines. From $$A$$, draw a line perpendicular to side $$BC$$. This line forms the altitude of the triangle, which we'll denote as $$AD$$. The length of $$AD$$ is equal to the vertical distance of $$A$$ from the other line, which is 2 units.

## Step 3: Understanding Equilateral Triangles

In an equilateral triangle, the altitude bisects the base forming two right triangles. It splits the triangle into two 30-60-90 special right triangles. Thus, the base $$BC$$ of the triangle is split into two equal lengths by the altitude $$AD$$. Let's denote one half of the base as $$BD$$.

## Step 4: Applying the Property of 30-60-90 Triangles

In a 30-60-90 triangle like $$ABD$$, the shorter leg $$BD$$ is half the hypotenuse $$AB$$ and the longer leg $$AD$$ is $$\sqrt{3}/2$$ times the hypotenuse $$AB$$. Using this property, we can find the length $$AB$$ by multiplying $$AD$$ by $$\frac{2}{\sqrt{3}}$$. So, $$AB = AD \times \frac{2}{\sqrt{3}} = 2 \times \frac{2}{\sqrt{3}}$$. Therefore, the length of a side of the equilateral triangle is $$\frac{4\sqrt{3}}{\sqrt{3}} = \sqrt{\frac{88}{3}}$$.

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