An isosceles triangle with two sides $\tau = 2\pi$ (one is the base), altitude 1, and area $\pi$. What is the length of third side and the three angles (in both radians and degrees)?

Note: not a right triangle, so values are not the same as Archimedes's triangle for the area of a disk.

  • 2
    $\begingroup$ What have you tried? You have the area. Imagine bisecting the angle between the two $2\pi$ sides to get two right triangles. $\endgroup$ May 19 '20 at 23:08


  • Find $AD$ by Pythagoras
  • Find $DC$ by subtraction
  • Find $BC$ by Pythagoras
  • Find angles by trigonometry

enter image description here

  • $\begingroup$ how did you draw that $\endgroup$ May 20 '20 at 3:10
  • $\begingroup$ is this geogebra $\endgroup$ May 20 '20 at 3:10
  • $\begingroup$ @endgameendgame: MSPaint - it is not to scale $\endgroup$
    – Henry
    May 20 '20 at 7:55

Say your triangle is $ABC$ with $AB=AC=\tau$ and the altitude $BH=1$, where $H$ in $AC$. You can use pythagorean theorem to get $AH=\sqrt{AB^2-BH^2}=\sqrt{\tau^2-1^2}$. Then, $HC=AC-AH=\tau-\sqrt{\tau^2-1^2}$. Finally, use pythagorean theorem again to compute that $$BC=\sqrt{BH^2+HC^2}=\sqrt{1+(\tau-\sqrt{\tau^2-1^2})^2}$$.

To obtain the angles, you express the angles in terms of inverse trig functions, for instance $\angle A=\arcsin(BH/AB)=\arcsin(1/\tau)$, and so on.

  • $\begingroup$ The last two angles can be done using: $\angle B = \angle C$ and $ (\pi - \arcsin(1/\tau))/2$. Thanks. Only one real use of $\pi$ nice and clean with $\tau$. $\endgroup$ May 20 '20 at 0:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.