1

I have a line whose length is "a". Can anyone explain procedure of drawing following lines?

1-line whose length is "a^2"

2-line whose length is "sqrt(a)"

Yaser
  • 31
  • 1
    Do you also have a line of length 1? –  Apr 15 '18 at 05:18
  • I've marked this as a duplicate for the construction of the square of a length. See my solution there. (As @martycohen remarks in his answer below: It's all in Euclid.) For the square root, you can use the same construction, simply replacing $x^2$ with $x$, and $x$ with $\sqrt{x}$. – Blue Apr 15 '18 at 06:14

1 Answers1

1

As Rahul wrote, you need to have a line of length 1.

Once you have this, you can use similar triangles to solve $\dfrac{1}{a} =\dfrac{a}{x} $ so that $x = a^2$.

To find $\sqrt{a}$, draw a circle with diameter $1+a$ and erect a perpendicular to a diameter a distance $1$ from the end. Where this intersects the circle, it makes a right triangle with the diameter, and the altitude $x$ satisfies $\dfrac{1}{x} =\dfrac{x}{a} $ so that $x^2 = a$ or $x = \sqrt{a}$.

Nothing original here - it's all in Euclid.

marty cohen
  • 110,450