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Problem 613

# Given the unit length below, construct a segment of length $$\sqrt{3}$$

Expert verified
Construct an equilateral triangle with side length 1, and then another adjacent equilateral triangle with side length 2. Split the larger triangle into two smaller equilateral triangles by drawing a perpendicular from one vertex to the opposite side. Apply the Pythagorean theorem to find the length of the altitude of the larger triangle, which is $$\sqrt{3}$$.
See the step by step solution

## Step 1: Construct an equilateral triangle of side length 1

Using the unit length given, construct an equilateral triangle ABC with each side equal to 1. To do this, you can use a compass to draw two circles with radii equal to the unit length, locating their centers on points A and B. The point where the two circles intersect will be C, the third vertex of the triangle.

## Step 2: Construct another equilateral triangle of side length 2

Extend one side of the triangle, say, segment AB. Using the unit length, draw a segment BD that is twice the length of AB. Construct an equilateral triangle BDE using the same method from step 1. You should now have two adjacent equilateral triangles of side lengths 1 and 2.

## Step 3: Split the larger triangle into two smaller equilateral triangles

Draw a line segment CE from vertex C to the opposite side of the larger triangle, DE, so that CE is perpendicular to DE. This forms two smaller equilateral triangles, CDE and BCE, with CD = DE = 1.

## Step 4: Use the Pythagorean theorem to find the length of CE

Recall the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. For triangle CDE, the hypotenuse is CB, which has a length of 2, and the other two sides are CD and CE. Using the Pythagorean theorem, we get: $CB^2 = CD^2 + CE^2$ $2^2 = 1^2 + CE^2$ $4 = 1 + CE^2$ $CE^2 = 3$ $CE = \sqrt{3}$

## Step 5: Construct the segment of length $$\sqrt{3}$$

Since CE is equal to $$\sqrt{3}$$, you can use the already constructed segment CE as the desired segment of length $$\sqrt{3}$$. In conclusion, you have now constructed a segment of length $$\sqrt{3}$$ using the Pythagorean theorem and the properties of equilateral triangles.

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