I am a blind high school student trying to make sense of all the geometric formulas that I have been told by my teacher this year. Currently, I have been trying to prove the formula for the surface area of a square pyramid.
My teacher has never proved to me why the surface area of a square pyramid is
$$S_A=\frac{PL}{2}+S^2$$
where $L$ is the slant height, $P$ is the perimeter of the base, and $S$ is the sidelength of the square base. I have decided to prove this formula myself. Here is my attempt:
Let $A, B, C, D$ denote the vertices of the base of the pyramid, and $T$ denote the apex, where the four slant sides converge into a single point. Because the base is a square,
$$AB=BC=CD=DA$$
It is also obvious that the latteral area of this square pyramid is the sum of all the areas of the triangles formed by the side length of the square base and the two slant heights respectively. Additionally, all slant heights have the same measure. Therefore, all four triangles composing the latteral surface area are congruent, and are also isosceles.
I also know that the area for a triangle is
$$A=\frac{BH}{2}$$
To find the height of the latteral-surface triangle, I can divide one of the latteral-surface triangles into two right triangles: by drawing an altitude from point $T$ to the base. The height of the two triangles created by the altitude are congruent, and the hypotenuse of each right triangle is also congruent. Therefore, by the Hypotenuse-leg law, I can conclude that the two triangles formed by the altitude are congruent. Thus, the point where the altitude intersects the base is the midpoint of the length of the square side. Then, I can use the Pythagorean law to find the height of the latteral-triangle:
$$(\frac{\text{length of the square base}}{2})^2+(\text{height of the latteral-triangle})^2=(\text{slant height of the pyramid})^2$$
Solving for the height of the latteral-surface triangle, I obtain
$$\text{height of the latteral-triangle}=\sqrt{\text{the slant height of the pyramid}^2-\frac{\text{side length of the square}}{2}^2}$$
With these facts established, I can write an equation to calculate the surface area of the square pyramid:
$$\frac{1}{2}(ab(\sqrt{at^2-\frac{(ab)^2}{4}}))+\frac{1}{2}(bc(\sqrt{bt^2-\frac{(bc)^2}{4}}))+\frac{1}{2}(cd(\sqrt{ct^2-\frac{(cd)^2}{4}}+\frac{1}{2}(da(\sqrt{dt^2-\frac{(da)^2}{4}}))+s^2$$
From this formula, I can factor out $\frac{1}{2}$. Additionally,
I have previously established that
$$ab=bc=cd=da$$. Therefore, I can replace $bc,cd,da$ with $ab$. Additionally, all of the slant heights are congruent, therefore, I can replace $at,bt,ct,dt$ with $at$.
The new equation then becomes
$$\frac{1}{2}(ab\sqrt{at^2-\frac{(ab)^2}{4}}+ab\sqrt{at^2-\frac{(ab)^2}{4}}+ab\sqrt{at^2-\frac{(ab)^2}{4}}+ab\sqrt{at^2-\frac{(ab)^2}{4}}+s^2$$
Since every term has $ab$ attached to it, I can factor it out. Also, I can combine the terms inside the parenthesis to get
$$\frac{ab}{2}(4\sqrt{at^2-\frac{ab^2}{4}})+s^2$$
which simplifies to
$$2ab\sqrt{at^2-\frac{ab^2}{4}}+s^2$$
This expression cannot be simplified any further, so that is the formula for the surface area. To test if my formula actually works, imagine eight 3-4-5 right triangles. If we pick up two of them and create an isosceles triangle such that the measures of this figure has a height of 4units, , a base of 6 units, and two 5 units sides. If we repeat this process four times, we will have four congruent triangles. Then if the four triangles are placed in such an angle where it forms a square pyramid.
When I used my formula to find the surface area of the pyramid, I get
$$2\cdot6\sqrt{5^2-\frac{(6)^2}{4}}+6^2$$
This expression yields $84$ square units.
However, if I use the formula that my teacher has provided me with, I obtain
$$\frac{6*4*5}{2}+36=96\text{square units}$$
What am I doing wrong?
Sorry if there is no visual aid, I have tried my best to describe the mental picture to you.
Best Answer
The slant height $L$ is the distance from the apex of the pyramid (the point $T$) to the nearest point on the perimeter of the base. This nearest point is the midpoint of any of the four edges of the square, and the distance $L$ in your example is $4$.
Your teacher’s formula then says $P=24$, $L=4$, and $S=6$, therefore
$$ \frac{PL}{2} + S^2 = \frac{24\times 4}{2}+6^2 = 84, $$
the same answer you calculated by your own method.
So congratulations, you have derived a correct formula for the area of the pyramid (it is indeed correct, it was not just due to luck that you got that answer to your example problem).
The visualization was excellent as well.