# 1/3 area of isoceles triangle at height “a” with given base b and height h

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!

• What is your question? – user061703 May 31 '18 at 3:11
• 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. – Joey B. May 31 '18 at 3:22
• So essentially the triangle above $a$ is $\frac 23 \frac 12bh$? – Tony Hellmuth May 31 '18 at 3:37
• Yes. The upper, similar triangle's area is 2/3 *1/2*bh – Joey B. May 31 '18 at 3:51
• This isn't too good of a "recreational-math" question. It feels more like a touch high-school geometry exercise. – 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 $\frac{\sqrt2}{\sqrt3}$. $$\frac{h-a}{h} = \frac{\sqrt2}{\sqrt3}$$ $$h - a = \frac{\sqrt{2}\cdot h}{\sqrt3}$$ $$a = h - \frac{\sqrt{2}\cdot h}{\sqrt3}$$
$$a = h\left (1 - \frac{\sqrt2}{\sqrt3} \right)$$
• Tip: \sqrt{2}. Or use \sqrt2 if the operand is single-digit. – Frenzy Li May 31 '18 at 5:57