Question

Use straightedge and compass to construct a line segment of length$\mathbf{1}\mathbf{+}\sqrt{\mathbf{3}}$ , beginning with the unit segment.

Short answer

$\sqrt{2}$is the hypotenuse of the right triangle having both sides $1$ .

$\sqrt{3}$is the hypotenuse of the right triangle having sides $1$ and$\sqrt{2}$ .

Step-by-step solution

Construct the circle.

Finding the length of BE.

Join the coordinates $\left(1,0\right)$and $\left(0,1\right)$, we get line segment AC .

Then , by Pythagoras’ theorem

$A{C}^2=A{B}^2+B{C}^2$where$AB=BC=1$

$$\begin{gathered} A{C}^2=\sqrt{1^2+1^2} \\ AC=\sqrt{2} \end{gathered}$$

Now, draw a circle with center at $\left(0,0\right)$ and radius $\sqrt{2}$. Assume the point of intersection with positive y-axis be D , then join DC .

Again , by pythagoras’ theorem

$D{C}^2=B{D}^2+B{C}^2$, we get

$DC=\sqrt{3}$

Now draw a circle with C as a center and radius $\sqrt{3}$. Suppose its point of intersection with positive x-axis be E , then $CE=\sqrt{3}$. As, CD and CE are radii of same circle.

$BE=BC+CE=1+\sqrt{3}$