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

From the volume formula: \(48=\dfrac13 S_{\text{base}}\cdot 4\Rightarrow S_{\text{base}}=36\text{ cm}^2\).

So the side of the square base is \(a=6\text{ cm}\).

Let \(O\) be the center of the base and \(M\) the midpoint of a side. Then \(OM=3\text{ cm}\), \(SO=4\text{ cm}\).

The slant height is \(SM=\sqrt{3^2+4^2}=5\text{ cm}\).

The area of one lateral face is \(\dfrac12\cdot 6\cdot 5=15\text{ cm}^2\).

The lateral surface area is \(4\cdot 15=60\text{ cm}^2\).

The total surface area is \(36+60=96\text{ cm}^2\).

Answer: \(96\text{ cm}^2\).

In a regular hexagon, the diagonal \(AC\) joining every second vertex equals \(a\sqrt3=4\sqrt3\text{ cm}\).

The circumradius of a regular hexagon equals its side: \(R=4\text{ cm}\).

The distance from the center of the base to diagonal \(AC\) is \(R\cos60^\circ=2\text{ cm}\).

The altitude of triangle \(SAC\) to \(AC\) is \(\sqrt{(2\sqrt2)^2+2^2}=2\sqrt3\text{ cm}\).

Hence the area of the section is \(\dfrac12\cdot 4\sqrt3\cdot 2\sqrt3=12\text{ cm}^2\).

Answer: \(12\text{ cm}^2\).

The base is an equilateral triangle with side \(6\sqrt3\text{ cm}\).

The circumradius of such a triangle is \(R=\dfrac{a}{\sqrt3}=6\text{ cm}\).

Let \(O\) be the center of the base. Then in the right triangle \(SAO\), the angle between edge \(SA\) and the base plane is \(\angle SAO=60^\circ\).

Hence \(\tan60^\circ=\dfrac{SO}{AO}\Rightarrow SO=6\sqrt3\text{ cm}\).

The base area is \(\dfrac{a^2\sqrt3}{4}=27\sqrt3\text{ cm}^2\).

The volume is \(\dfrac13\cdot 27\sqrt3\cdot 6\sqrt3=162\text{ cm}^3\).

Answer: \(162\text{ cm}^3\).

Let \(O\) be the center of the base. Then \(AO=\sqrt{15^2-12^2}=9\text{ cm}\).

For a square, \(AO=\dfrac{a\sqrt2}{2}\), so \(a=9\sqrt2\text{ cm}\).

The base area is \((9\sqrt2)^2=162\text{ cm}^2\).

The volume of the whole pyramid is \(\dfrac13\cdot 162\cdot 12=648\text{ cm}^3\).

Since \(AK=5\text{ cm}\), we have \(SK=10\text{ cm}\). The small top pyramid is similar to the original one with coefficient \(\dfrac{SK}{SA}=\dfrac{10}{15}=\dfrac23\).

Its volume is \(648\cdot \left(\dfrac23\right)^3=192\text{ cm}^3\).

The volume of the frustum is \(648-192=456\text{ cm}^3\).

Answer: \(456\text{ cm}^3\).

If all lateral faces have the same inclination angle, the projection of the apex is the incenter of the base.

The altitude to the base \(12\) in the triangle with sides \(10,10,12\) is \(\sqrt{10^2-6^2}=8\text{ cm}\).

The base area is \(\dfrac{12\cdot 8}{2}=48\text{ cm}^2\).

The semiperimeter is \(\dfrac{10+10+12}{2}=16\text{ cm}\).

The inradius is \(r=\dfrac{48}{16}=3\text{ cm}\).

Since \(\tan45^\circ=1\), the pyramid height equals \(r\), i.e. \(3\text{ cm}\).

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

The area of the rhombus is \(\dfrac{4\cdot 6}{2}=12\text{ cm}^2\).

The intersection point of the diagonals is at distances equal to half the diagonals from the vertices: \(2\text{ cm}\) and \(3\text{ cm}\).

The longer lateral edge corresponds to the larger distance, i.e. \(3\text{ cm}\).

Let the height of the pyramid be \(h\). Then \(h^2+3^2=5^2\Rightarrow h=4\text{ cm}\).

The volume is \(\dfrac13\cdot 12\cdot 4=16\text{ cm}^3\).

Answer: \(16\text{ cm}^3\).

The base is a right triangle with sides \(6,8,10\).

Its area is \(\dfrac{6\cdot 8}{2}=24\text{ cm}^2\), and its semiperimeter is \(\dfrac{6+8+10}{2}=12\text{ cm}\).

The inradius of the base is \(r=\dfrac{24}{12}=2\text{ cm}\).

With the same inclination angle for all lateral faces, the projection of the apex is the incenter of the base.

From the right triangle perpendicular to a base side, \(\tan60^\circ=\dfrac{h}{r}\), hence \(h=2\sqrt3\text{ cm}\).

The slant height of each lateral face is \(l=\sqrt{h^2+r^2}=\sqrt{12+4}=4\text{ cm}\).

Therefore, the lateral area is \(\dfrac12\cdot P_{\text{base}}\cdot l=\dfrac12\cdot 24\cdot 4=48\text{ cm}^2\).

Answer: \(48\text{ cm}^2\).

The base area is \(\dfrac{16\cdot 8}{2}=64\text{ cm}^2\).

Since \(SC\perp(ABC)\), the height of the pyramid is \(12\text{ cm}\).

The volume of the whole pyramid is \(\dfrac13\cdot 64\cdot 12=256\text{ cm}^3\).

The small top pyramid is similar to the large one with coefficient \(\dfrac{SK}{SC}=\dfrac{3}{12}=\dfrac14\).

Its volume is \(256\cdot \left(\dfrac14\right)^3=4\text{ cm}^3\).

The volume of the frustum is \(256-4=252\text{ cm}^3\).

Answer: \(252\text{ cm}^3\).

The area of the rhombus is \(a^2\sin60^\circ=16\cdot \dfrac{\sqrt3}{2}=8\sqrt3\text{ cm}^2\).

Let the foot of the height be the acute-angle vertex \(A\). The farthest vertex from it is the opposite vertex \(C\), so the longer lateral edge is \(SC\).

The larger diagonal of a rhombus with side \(4\) and angle \(60^\circ\) is \(4\sqrt3\text{ cm}\).

Then in the right triangle \(SAC\):

\(h^2+(4\sqrt3)^2=(5\sqrt3)^2\Rightarrow h^2=27\Rightarrow h=3\sqrt3\text{ cm}\).

The volume is \(\dfrac13\cdot 8\sqrt3\cdot 3\sqrt3=24\text{ cm}^3\).

Answer: \(24\text{ cm}^3\).

The base is an equilateral triangle with side \(12\text{ cm}\), so its area is \(\dfrac{12^2\sqrt3}{4}=36\sqrt3\text{ cm}^2\).

Let \(O\) be the center of the base and \(M\) the midpoint of a side. Then \(OM\) is the inradius of the equilateral triangle: \(\dfrac{12\sqrt3}{6}=2\sqrt3\text{ cm}\).

In triangle \(SOM\), the angle \(\angle SMO=45^\circ\), hence \(SO=OM=2\sqrt3\text{ cm}\).

The volume of the pyramid is \(\dfrac13\cdot 36\sqrt3\cdot 2\sqrt3=72\text{ cm}^3\).

Answer: \(72\text{ cm}^3\).

The altitude of the base triangle to side \(AC\) is \(\sqrt{20^2-16^2}=12\text{ cm}\).

The base area is \(\dfrac12\cdot 32\cdot 12=192\text{ cm}^2\).

Since \(SB\perp(ABC)\), triangles \(SAB\) and \(SBC\) are right triangles, and each has area \(\dfrac12\cdot 20\cdot 9=90\text{ cm}^2\).

Now find face \(SAC\). We have \(SA=SC=\sqrt{20^2+9^2}=\sqrt{481}\text{ cm}\).

The altitude of triangle \(SAC\) to base \(AC\) is \(\sqrt{481-16^2}=15\text{ cm}\).

Therefore \(S_{SAC}=\dfrac12\cdot 32\cdot 15=240\text{ cm}^2\).

The total surface area is \(192+90+90+240=612\text{ cm}^2\).

Answer: \(612\text{ cm}^2\).