Pyramid Triangles

Question

A pyramid has a triangular base, length $300$ and has a height (center of the equilateral triangle to the vertex) of $100$. I climb the edge of the pyramid to the top and I have covered $138$ cm, what is my altitude?

I have tried using the approach of similar triangles, the center of the equilateral triangle connected to a vertex has length $50\sqrt3$, then the length of the edge must be $\sqrt{10000 + 75000} = 50$ $\sqrt7$. Using similar triangles, the ratio was - $50 \sqrt7 /100 = 138/\text{altitude}$ which is wrong.

from sipnayan 2017

Answer 1

Hint: find distance from vertex to centre of base.

Answer 2

The side is $VB=\sqrt{100^2+(100\sqrt{3})^2}=200$

because $BH=\frac23 h$

Answer 3

Now that you've clarified this is a past competition, I can help.

Consider the right triangle formed by height of pyramid (from apex of pyramid to base, which is the centre of the equilateral base) - which also constitutes "height" of this right triangle, the edge of the pyramid (the "hypotenuse" of this triangle) and the line segment connecting the vertex of the base to the center of the base (let's call that the "base" of the right triangle).

The last line segment ("base" of right triangle) is $\frac 23$ the length of the height of the base equilateral triangle (centroid theorem). The height of the base triangle is $300\cos 30^{\circ} = 150\sqrt 3$. So the length of the "base" is $\frac 23 \cdot 150\sqrt 3 = 100\sqrt 3$.

By Pythagoras' theorem, the "hypotenuse" of the right triangle (also the edge of the pyramid) is $\sqrt{(100\sqrt 3)^2 + 100^2} = 200$

Hence, by similar triangles, the altitude you require is $\frac{139}{200} \cdot 100 = 69.5$.