Since all lateral edges are equal, the projection of the apex onto the base is the center of the rectangle.

The diagonal of the base is \(\sqrt{6^2+8^2}=10\text{ cm}\), so the distance from the center to a vertex of the base is \(5\text{ cm}\).

Let the height of the pyramid be \(h\). Then

\(h^2+5^2=61\Rightarrow h=6\text{ cm}\).

The base area is \(6\cdot 8=48\text{ cm}^2\).

The volume is \(V=\dfrac13\cdot 48\cdot 6=96\text{ cm}^3\).

Answer: \(96\text{ cm}^3\).

If the diagonal of a square is \(10\sqrt2\), then its side is \(a=10\text{ cm}\).

The base area is \(a^2=100\text{ cm}^2\).

Let \(O\) be the center of the base and \(M\) the midpoint of a side. Then \(OM=\dfrac a2=5\text{ cm}\).

In triangle \(SOM\), the angle \(\angle SMO\) equals the angle between the lateral face and the base plane, i.e. \(60^\circ\).

\(\cos60^\circ=\dfrac{OM}{SM}=\dfrac5{SM}\Rightarrow SM=10\text{ cm}\).

This is the slant height of the pyramid. Area of one lateral face: \(\dfrac12\cdot 10\cdot 10=50\text{ cm}^2\).

The lateral area is \(4\cdot 50=200\text{ cm}^2\).

The total surface area is \(100+200=300\text{ cm}^2\).

Answer: \(300\text{ cm}^2\).

If all lateral faces are inclined to the base at the same angle, the projection of the apex onto the base is the incenter of the base. For a rhombus, this is the intersection point of its diagonals.

The area of the rhombus is \(S=\dfrac{40\cdot 30}{2}=600\text{ cm}^2\).

Its side is \(a=\sqrt{20^2+15^2}=25\text{ cm}\).

The semiperimeter is \(p=2a=50\text{ cm}\).

The inradius is \(r=\dfrac{S}{p}=\dfrac{600}{50}=12\text{ cm}\).

Since \(\tan45^\circ=1\), the height of the pyramid equals this distance: \(h=r=12\text{ cm}\).

Answer: \(12\text{ cm}\).

The side of the square base is \(a=\sqrt{32}=4\sqrt2\text{ cm}\).

Let \(O\) be the center of the base. Then \(OM=OK=\dfrac a2=2\sqrt2\text{ cm}\), and \(SO=2\sqrt3\text{ cm}\).

By the Pythagorean theorem,

\(SM=SK=\sqrt{SO^2+OM^2}=\sqrt{12+8}=2\sqrt5\text{ cm}\).

The segment \(MK\) joins the midpoints of adjacent sides of the square, so

\(MK=\sqrt{\left(\dfrac a2\right)^2+\left(\dfrac a2\right)^2}=4\text{ cm}\).

Triangle \(SMK\) is isosceles with sides \(2\sqrt5,2\sqrt5,4\). Its altitude to the base \(MK\) is \(\sqrt{(2\sqrt5)^2-2^2}=4\text{ cm}\).

The area of the section is \(\dfrac12\cdot 4\cdot 4=8\text{ cm}^2\).

Answer: \(8\text{ cm}^2\).

The hypotenuse of the right triangle base is \(\sqrt{10^2+24^2}=26\text{ cm}\).

If all lateral edges are equal, the projection of the apex is the circumcenter of the base. For a right triangle, this is the midpoint of the hypotenuse.

The circumradius is \(R=13\text{ cm}\).

Let the pyramid height be \(h\). Then

\(h^2+13^2=269\Rightarrow h=10\text{ cm}\).

The base area is \(\dfrac{10\cdot 24}{2}=120\text{ cm}^2\).

The volume is \(V=\dfrac13\cdot 120\cdot 10=400\text{ cm}^3\).

Answer: \(400\text{ cm}^3\).