I was thinking of interesting and difficult geometry problems to attempt to solve when I came up with this:

Given and isosceles triangle with given base b and height h, find at what height, a, the area under a is equal to one third the area of the original triangle, in terms of b and h. Here is a diagram of the problem. Hope that you find this interesting and come up with a solution in terms of b and h.

Have fun!

  • 1
    $\begingroup$ What is your question? $\endgroup$ – user061703 May 31 '18 at 3:11
  • $\begingroup$ Find an expression to represent at what height, a, is the area of the triangle under "a" 1/3 the area of the whole triangle. $\endgroup$ – Joey B. May 31 '18 at 3:22
  • $\begingroup$ So essentially the triangle above $a$ is $\frac 23 \frac 12bh$? $\endgroup$ – Tony Hellmuth May 31 '18 at 3:37
  • $\begingroup$ Yes. The upper, similar triangle's area is 2/3 *1/2*bh $\endgroup$ – Joey B. May 31 '18 at 3:51
  • $\begingroup$ This isn't too good of a "recreational-math" question. It feels more like a touch high-school geometry exercise. $\endgroup$ – Mike Pierce Jun 2 '18 at 20:35

The upper triangle is similar to the larger triangle and their area ratio is $\frac{2}{3}$. Their length ratio is therefore $\frac{\sqrt2}{\sqrt3}$. $$\frac{h-a}{h} = \frac{\sqrt2}{\sqrt3}$$ $$h - a = \frac{\sqrt{2}\cdot h}{\sqrt3}$$ $$a = h - \frac{\sqrt{2}\cdot h}{\sqrt3}$$

$$a = h\left (1 - \frac{\sqrt2}{\sqrt3} \right)$$

  • $\begingroup$ Tip: \sqrt{2}. Or use \sqrt2 if the operand is single-digit. $\endgroup$ – Frenzy Li May 31 '18 at 5:57

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.