Why the Formulas Work

Step 1 — a pentagon fans into $5 - 2 = 3$ triangles, so its angles total $540^\circ$.

Step 2 — the known angles add to $420^\circ$.

Step 3 — the fifth is $540 - 420 = 120^\circ$.

Step 1 — each exterior angle is $180 - 156 = 24^\circ$.

Step 2 — the exterior angles total $360^\circ$ (one full turn), so there are $360/24 = 15$ of them.

Step 3 — $15$ sides. The exterior-angle route avoids solving $(n - 2)180 = 156n$, though that equation gives the same $n = 15$.

Step 1 — the wall, the ground and the ladder form a right triangle with the ladder as hypotenuse.

Step 2 — $h^2 + 5^2 = 13^2$, so $h^2 = 169 - 25 = 144$.

Step 3 — $h = 12$ feet: the $5$-$12$-$13$ triple.

Step 1 — the longest rod runs corner to opposite corner.

Step 2 — the diagonal of the $3 \times 4$ end face is $\sqrt{9 + 16} = 5$.

Step 3 — that diagonal is perpendicular to the $12$ edge, so the rod is the hypotenuse of a right triangle with legs $5$ and $12$: $13$. In one line, $\sqrt{12^2 + 3^2 + 4^2} = \sqrt{169} = 13$.

Step 1 — count the terms: from $3$ to $99$ in steps of $4$ is $\tfrac{99 - 3}{4} = 24$ steps, so $25$ terms.

Step 2 — Gauss pairing: first plus last $= 3 + 99 = 102$, and every pair from the ends has that sum.

Step 3 — sum $= \tfrac{25\cdot102}{2} = 1275$.

Step 1 — do not compute either sum. Pair them up: $(2 - 1) + (4 - 3) + \cdots + (100 - 99)$.

Step 2 — there are $50$ pairs, each worth $1$.

Step 3 — the answer is $50$. (Check with the formulas: $2550 - 2500 = 50$.)

Copy the triangle, rotate the copy $180^\circ$ about the midpoint of one side, and the two together tile a parallelogram with the same base and height. A parallelogram of base $b$ and height $h$ is a rectangle after cutting one end triangle and sliding it to the other side, so it has area $bh$ — and the original triangle is exactly half of it. $\blacksquare$ The same sliding argument shows why the height must be measured PERPENDICULAR to the base, which is the step most often skipped.

Cut the disc into $2n$ equal sectors and lay them alternately point-up and point-down: the result is nearly a parallelogram of height $r$ and base half the circumference, $\pi r$. As $n$ grows the wiggly edges straighten, and the area tends to $\pi r\cdot r = \pi r^2$. $\blacksquare$ The same picture explains the surface-area-to-volume relation of a sphere, and it is the reason $\pi$ appears in both the circumference and the area.

Step 1 — take a second copy, turn it upside down, and place it beside the first: together they form a parallelogram with base $6 + 10 = 16$ and height $4$.

Step 2 — its area is $16\cdot4 = 64$.

Step 3 — the trapezoid is half: $32$. That is the formula $\tfrac{b_1 + b_2}{2}h$, rebuilt in three lines.

Step 1 — join the centre to the six vertices. The six triangles are identical (the hexagon is regular), so their angles at the centre are $360/6 = 60^\circ$ each; each interior angle of the hexagon is $120^\circ$ and is split in half, giving $60^\circ$ base angles. So all six triangles are equilateral with side $2$.

Step 2 — Pythagoras gives each triangle's height: $\sqrt{2^2 - 1^2} = \sqrt3$, so each has area $\tfrac12\cdot2\cdot\sqrt3 = \sqrt3$.

Step 3 — the hexagon has area $6\sqrt3$. Three proved formulas, one answer.