19. Kevin makes a net using a combination of squares and equilateral triangles, as shown in the figure. The side-length of each square and of each triangle is $1 \mathrm{~cm}$. He folds the net up to form the surface of a polyhedron. What is the distance between the vertices $J$ and $K$ in this polyhedron?
A $\sqrt{5} \mathrm{~cm}$
B $(1+\sqrt{2}) \mathrm{cm}$
C $\frac{5}{2} \mathrm{~cm}$
D $(1+\sqrt{3}) \mathrm{cm}$
E $2 \sqrt{2} \mathrm{~cm}$
Solution
B
The net is folded to form a polyhedron that consists of a cube with a square based pyramid on the top and bottom faces. $J$ becomes the apex of the pyramid on the top face, and $K$ the apex of the pyramid on the bottom face.
Therefore the distance between $J$ and $K$ equals the height of the cube plus the heights of the pyramids.
The 4 vertices joined to $J$ form a square $P Q R S$ of side $1 \mathrm{~cm}$ long. Therefore, by the Pythagorean theorem, its diagonal $S Q$ is $\sqrt{2} \mathrm{~cm}$ long. Let $M$ be the midpoint of $S Q$. Then $J M Q$ is a right triangle with hypotenuse $J Q$ of side $1 \mathrm{~cm}$ and leg $Q M$ of side $\frac{\sqrt{2}}{2} \mathrm{~cm}$. Again, using the Pythagorean theorem, we get
$$
|J M|^{2}=1-\left(\frac{\sqrt{2}}{2}\right)^{2}=1-\frac{2}{4}=\frac{1}{2},
$$
thus
$$
|J M|=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} .
$$

The distance between $J$ and $K$ is
$$
\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}=1+\sqrt{2} \mathrm{~cm}
$$