Proof that length is proportional to height

Question

For problem 4a. I saw the solution to be 3.6cm and it was worked out by the equation 5/8 = 6/(6+h). But why can we establish the above relationship? The length 6cm is not the height of the triangle with base 5cm, so I don't see how the two are related. Why is the side proportional to the height in a simple explanation?

Answer 1

Have a look at the theorem of intersecting lines. This basically tells you that the relation between the parts you described in your example are equal.

The intuition behind that is the following: The outer line is linearly decreasing the width of the underlying triangle from 8 cm down to 0 cm at the top. As this is done linearly, you know cutting it through the middle (at half of the height), the triangle will have exactly half the width (4 cm). Similarly, if the triangle has $5/8$ of the original width, the remaining 6 cm equal $5/8$ of the total height $h$.

In formulas: $6 = 5/8 \cdot h$, so $h=6 \cdot 8/5=9.6$ so the remaining bottom part has a height of $9.6-6=3.6$ cm.

Answer 2

If we say the top triangular pyramid has the closest face called triangle A, we know it has a base of width 5cm and height of 6cm.

Imagine that the top is still on the truncated piece, then we know that triangle A must be a similar triangle to the combined face. We know they are similar triangles because all the angles must be the same (which can be proven from knowing the cut was made parallel). Hence, the size ratio for each edge must remain constant due to similar edges.

Accordingly, if you say S is the size of A and C is the size of the combined shape:

Ratio = $ \frac {S}{C} = \frac {5}{8} = \frac {6}{6+h}$

Then rearranging, you end up with h = 3.6

I hope this helps.

Answer 3

Place the origin of the Cartesian coordinate frame at the apex (the top of the pyramid). Now, let the three edges be along the unit vectors $\vec{v_1}, \vec{v_2}, \vec{v_3}$, and let the vertical axis of the pyramid be along the unit vector $\vec{a}$.

Suppose you cut the pyramid perpendicular to its axis by a cutting plane, then this cutting plane has the equation

$$ \vec{a} \cdot ( \vec{r} - t \vec{a} ) = 0 $$

i.e.

$$ \vec{a} \cdot \vec{r} = t $$

where $r = (x,y,z) $

To find where this cutting plane cuts through the edges, let $\vec{r} = s_1 \vec{v_1}$, then

$ \vec{a} \cdot (s_1 \vec{v_1}) = t $

Hence

$ s_1 = \bigg(\dfrac{1}{\vec{a} \cdot \vec{v_1}} \bigg) t = k_1 t$

similarly, for the second edge

$ s_2 = \bigg( \dfrac{1}{\vec{a} \cdot \vec{v_2}} \bigg) t = k_2 t $

The side length of the triangle of intersection between these two edges is the length of the vector

$ \vec{s} = s_2 \vec{v_2} - s_1 \vec{v_1} = t \bigg( k_2 \vec{v_2} - k_1 \vec{v_1} \bigg) $

Hence, the length of this vector is

$ \| \vec{s} \| = t \| k_2 \vec{v_2} - k_1 \vec{v_1} \| = K t $

This establishes the linear relation between the length of the side and the distance $ t$ along the axis.

Answer 4

Place the apex at the origin, and scale the small pyramid by a factor of 8/5 (centred at origin). Then its base, which was originally 6 cm away from the origin, becomes 6 cm · 8/5 away from the origin after the scaling.