Prisms and Pyramids Problems: Difficult Problems with Solutions

Solution:
Let the frustum be [tex]ABCDA_1B_1C_1D_1[/tex] with bases the squares [tex]ABCD[/tex], [tex]AB=7[/tex], and [tex]A_1B_1C_1D_1[/tex], [tex]A_1B_1=5[/tex], and let [tex]O[/tex] and [tex]O_1[/tex] be the centers of the bases; then [tex]OO_1[/tex] is the height of the frustum. The diagonal section [tex]ACC_1A_1[/tex] contains [tex]OO_1[/tex], so it is perpendicular to the base; it is an isosceles trapezoid with bases
[tex]AC=7\sqrt{2},\quad A_1C_1=5\sqrt{2}[/tex]
Let [tex]H[/tex] be the foot of the perpendicular from [tex]A_1[/tex] to [tex]AC[/tex]. The segment [tex]A_1H[/tex] lies in a plane perpendicular to the base and is perpendicular to the common line [tex]AC[/tex], so [tex]A_1H\perp(ABC)[/tex] and [tex]A_1H=h[/tex] is the height of the frustum. In the isosceles trapezoid
[tex]AH=\frac{AC-A_1C_1}{2}=\sqrt{2},\quad HC=AC-AH=6\sqrt{2}[/tex]
From the right triangle [tex]A_1HC[/tex], whose hypotenuse is the diagonal [tex]A_1C=9[/tex] of the frustum:
[tex]h^2=A_1C^2-HC^2=81-72=9\Rightarrow h=3[/tex]
The base areas are [tex]S_1=7^2=49[/tex] and [tex]S_2=5^2=25[/tex], so by the formula for the volume of a frustum
[tex]V=\frac{h}{3}\left(S_1+S_2+\sqrt{S_1S_2}\right)=\frac{3}{3}(49+25+35)=109[/tex]

Solution:
Let [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] be the edges of the container in inches. When the container stands on a face, the water fills a rectangular box with that face as its base and the water level as its height, and its volume is [tex]90\,000[/tex] cubic inches. Therefore
[tex]ab\times 30=90\,000,\quad ac\times 37.5=90\,000,\quad bc\times 45=90\,000[/tex]
[tex]ab=3000,\quad ac=2400,\quad bc=2000[/tex]
Multiplying the three equalities:
[tex](abc)^2=3000\times 2400\times 2000=144\times 10^8\Rightarrow V=abc=12\times 10^4=120\,000[/tex]
So the volume of the container is [tex]120\,000[/tex] cubic inches, i.e. [tex]120[/tex] thousand cubic inches. Check: the edges are [tex]a=\frac{abc}{bc}=60[/tex], [tex]b=\frac{abc}{ac}=50[/tex], [tex]c=\frac{abc}{ab}=40[/tex], and in each position the water level is below the top of the container: [tex]30\lt 40[/tex], [tex]37.5\lt 50[/tex], [tex]45\lt 60[/tex].

Solution:
Let [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] be the edges from one vertex. If the face with sides [tex]a[/tex] and [tex]b[/tex] is the base, the lateral faces are two rectangles with sides [tex]a[/tex], [tex]c[/tex] and two with sides [tex]b[/tex], [tex]c[/tex], so the lateral surface area is [tex]2ac+2bc[/tex]. For the other two choices of the base it is [tex]2ab+2ac[/tex] and [tex]2ab+2bc[/tex].
Put [tex]x=ab[/tex], [tex]y=bc[/tex], [tex]z=ca[/tex]. Naming the edges suitably, we may assume that
[tex]2(y+z)=10, \quad 2(x+z)=16, \quad 2(x+y)=18\Rightarrow y+z=5, \quad x+z=8, \quad x+y=9[/tex]
Adding the three equations gives [tex]2(x+y+z)=22[/tex], i.e. [tex]x+y+z=11[/tex], hence
[tex]x=6, \quad y=3, \quad z=2[/tex]
For the volume [tex]V=abc[/tex] we have
[tex]V^2=ab\times bc\times ca=xyz=36\Rightarrow V=6[/tex]
Check: the edges are [tex]a=2[/tex], [tex]b=3[/tex], [tex]c=1[/tex], and the three lateral surface areas are [tex]2\times 1\times (2+3)=10[/tex], [tex]2\times 2\times (3+1)=16[/tex] and [tex]2\times 3\times (1+2)=18[/tex].

Solution:
Let [tex]ABCM[/tex] be the pyramid with apex [tex]M[/tex], [tex]MA=MB=MC=10[/tex] and base edge [tex]AB=2x[/tex]. The three lateral faces are congruent isosceles triangles, so each of them has area [tex]\frac{144}{3}=48[/tex].
Let [tex]K[/tex] be the midpoint of [tex]AB[/tex]. In the isosceles triangle [tex]ABM[/tex] the median [tex]MK[/tex] is also the altitude (the slant height), and from the right triangle [tex]AKM[/tex] we get [tex]MK=\sqrt{100-x^2}[/tex]. The area of the face gives
[tex]\frac{1}{2}\times 2x\times \sqrt{100-x^2}=48\Rightarrow x^2(100-x^2)=2304\Rightarrow x^4-100x^2+2304=0[/tex]
This biquadratic equation gives [tex]x^2=36[/tex] or [tex]x^2=64[/tex], i.e. [tex]x=6[/tex] or [tex]x=8[/tex], so [tex]AB=12[/tex] or [tex]AB=16[/tex].
Both pyramids exist: the foot of the height [tex]h[/tex] is the circumcenter of the base, whose circumradius is [tex]R=\frac{AB}{\sqrt{3}}[/tex], so [tex]h^2=MA^2-R^2=100-\frac{AB^2}{3}[/tex]; for [tex]AB=12[/tex] this gives [tex]h^2=52\gt 0[/tex], and for [tex]AB=16[/tex] it gives [tex]h^2=\frac{44}{3}\gt 0[/tex].
Therefore the base edge is [tex]12[/tex] in or [tex]16[/tex] in.

Solution:
Let [tex]ABCDA_1B_1C_1D_1[/tex] be the truncated pyramid, [tex]AB=2[/tex], [tex]A_1B_1=1[/tex], and let [tex]O[/tex], [tex]O_1[/tex] be the centers of the bases, [tex]OO_1=3[/tex]. The diagonals [tex]AC_1[/tex] and [tex]A_1C[/tex] lie in the diagonal section [tex]ACC_1A_1[/tex], an isosceles trapezoid with bases [tex]AC=2\sqrt{2}[/tex] and [tex]A_1C_1=\sqrt{2}[/tex]; let them meet at [tex]T[/tex]. By symmetry all four diagonals pass through this point of the axis [tex]OO_1[/tex].
The triangles [tex]TAC[/tex] and [tex]TC_1A_1[/tex] are similar ([tex]AC\parallel A_1C_1[/tex]) with ratio [tex]\frac{AC}{A_1C_1}=2[/tex], so their heights from [tex]T[/tex], i.e. the distances from [tex]T[/tex] to the two bases, are in ratio [tex]2:1[/tex]. Hence [tex]T[/tex] is at distance [tex]2[/tex] from the larger base and [tex]1[/tex] from the smaller one.
Extend the lateral edges to the full pyramid with apex [tex]M[/tex]. Since the base edges are in ratio [tex]1:2[/tex], the full pyramid has height [tex]6[/tex]. The section through [tex]T[/tex] is a square at distance [tex]6-2=4[/tex] from [tex]M[/tex], so its side [tex]x[/tex] and its area [tex]S[/tex] are
[tex]x=2\times \frac{4}{6}=\frac{4}{3}\Rightarrow S=x^2=\frac{16}{9}[/tex]
By the formula [tex]V=\frac{h}{3}\left(S_1+S_2+\sqrt{S_1S_2}\right)[/tex] for the two smaller truncated pyramids (with heights [tex]2[/tex] and [tex]1[/tex]):
[tex]V_1=\frac{2}{3}\left(4+\frac{16}{9}+\frac{8}{3}\right)=\frac{152}{27}, \quad V_2=\frac{1}{3}\left(\frac{16}{9}+1+\frac{4}{3}\right)=\frac{37}{27}[/tex]
Check: [tex]V_1+V_2=7=\frac{3}{3}(4+1+2)[/tex], the volume of the whole truncated pyramid.

Solution:
Let [tex]MO[/tex] be the height of the pyramid and [tex]K[/tex], [tex]L[/tex], [tex]N[/tex] the feet of the perpendiculars from [tex]O[/tex] to the lines [tex]AB[/tex], [tex]BC[/tex], [tex]CA[/tex]. By the theorem of three perpendiculars [tex]MK\perp AB[/tex], [tex]ML\perp BC[/tex], [tex]MN\perp CA[/tex], so [tex]\angle MKO[/tex], [tex]\angle MLO[/tex], [tex]\angle MNO[/tex] are the linear angles of the dihedral angles at the base edges and each of them equals [tex]45^{\circ}[/tex].
The right triangles [tex]MOK[/tex], [tex]MOL[/tex], [tex]MON[/tex] have the common leg [tex]MO[/tex] and equal acute angles, so they are congruent and [tex]OK=OL=ON[/tex]. The dihedral angles are acute, so [tex]O[/tex] lies inside [tex]\triangle ABC[/tex]; hence [tex]O[/tex] is the incenter of [tex]\triangle ABC[/tex] and [tex]OK=r[/tex] is its inradius. The right triangle [tex]MOK[/tex] has a [tex]45^{\circ}[/tex] angle, so it is isosceles and [tex]MO=OK=r[/tex].
In the isosceles triangle [tex]ABC[/tex] the height [tex]CH[/tex] is also a median, so [tex]AH=3[/tex], and from the right triangle [tex]AHC[/tex]:
[tex]CH=\sqrt{5^2-3^2}=4,\quad S_{ABC}=\frac{6\times 4}{2}=12[/tex]
The semiperimeter is [tex]p=\frac{6+5+5}{2}=8[/tex], hence
[tex]MO=r=\frac{S_{ABC}}{p}=\frac{12}{8}=\frac{3}{2}[/tex]
The volume of the pyramid is
[tex]V=\frac{1}{3}S_{ABC}\times MO=\frac{1}{3}\times 12\times \frac{3}{2}=6[/tex]

Solution:
Let [tex]a[/tex] be the base edge, [tex]l[/tex] the lateral edge, [tex]K[/tex] the midpoint of [tex]AB[/tex] and [tex]k=MK[/tex] the slant height (apothem). The three lateral faces are congruent isosceles triangles, so the total surface area is [tex]S=S_{ABC}+3S_{ABM}[/tex]. Since [tex]S_{ABC}=\frac{S}{4}[/tex],
[tex]3S_{ABM}=S-S_{ABC}=\frac{3S}{4}=3S_{ABC}\Rightarrow S_{ABM}=S_{ABC}[/tex]
In the isosceles face [tex]ABM[/tex] the median [tex]MK[/tex] is also a height, so
[tex]\frac{1}{2}ak=\frac{a^2\sqrt{3}}{4}\Rightarrow k=\frac{a\sqrt{3}}{2}[/tex]
From the right triangle [tex]MKB[/tex]:
[tex]l^2=k^2+\left(\frac{a}{2}\right)^2=\frac{3a^2}{4}+\frac{a^2}{4}=a^2\Rightarrow l=a[/tex]
Thus all edges of the pyramid are equal (it is a regular tetrahedron). The pyramid has [tex]3[/tex] base edges and [tex]3[/tex] lateral edges, so
[tex]6a=120\Rightarrow l=a=20[/tex]

Solution:
Since [tex]A_1C\perp(ABC)[/tex], the segment [tex]A_1C=h[/tex] is the height of the prism. Draw [tex]CK\perp AB[/tex] ([tex]K\in AB[/tex]). By the theorem of three perpendiculars [tex]A_1K\perp AB[/tex], so [tex]\angle A_1KC=60^{\circ}[/tex] is the linear angle of the dihedral angle at [tex]AB[/tex]. Put [tex]CK=x[/tex]; from the right triangle [tex]A_1CK[/tex] we get [tex]h=x\sqrt{3}[/tex] and [tex]A_1K=2x[/tex].
The area of the base is [tex]S_{ABC}=\frac{AB\times CK}{2}=\frac{3x}{2}[/tex], so
[tex]V=\frac{3x}{2}\times x\sqrt{3}=\frac{3\sqrt{3}}{2}x^2=3\sqrt{3}\Rightarrow x=\sqrt{2},\ h=\sqrt{6},\ A_1K=2\sqrt{2}[/tex]
In the right triangle [tex]ABC[/tex] we have [tex]AC^2+BC^2=AB^2=9[/tex] and [tex]AC\times BC=AB\times CK=3\sqrt{2}[/tex], so
[tex](AC+BC)^2=9+6\sqrt{2}=3(\sqrt{2}+1)^2,\quad (AC-BC)^2=9-6\sqrt{2}=3(\sqrt{2}-1)^2[/tex]
Since [tex]AC\gt BC[/tex], we get [tex]AC+BC=\sqrt{3}(\sqrt{2}+1)[/tex] and [tex]AC-BC=\sqrt{3}(\sqrt{2}-1)[/tex], hence [tex]AC=\sqrt{6}[/tex], [tex]BC=\sqrt{3}[/tex] and each base has area [tex]S_{ABC}=\frac{\sqrt{6}\times \sqrt{3}}{2}=\frac{3\sqrt{2}}{2}[/tex].
The face [tex]ABB_1A_1[/tex] is a parallelogram with side [tex]AB=3[/tex] and corresponding height [tex]A_1K=2\sqrt{2}[/tex], so [tex]S_{ABB_1A_1}=3\times 2\sqrt{2}=6\sqrt{2}[/tex].
In the parallelogram [tex]ACC_1A_1[/tex] we have [tex]A_1C\perp AC[/tex], so [tex]A_1C[/tex] is its height to the side [tex]AC[/tex] and [tex]S_{ACC_1A_1}=AC\times A_1C=\sqrt{6}\times \sqrt{6}=6[/tex].
Finally, [tex]BC\perp AC[/tex] and [tex]BC\perp A_1C[/tex], so [tex]BC\perp(ACC_1A_1)[/tex] and [tex]BC\perp CC_1[/tex]. Thus [tex]BCC_1B_1[/tex] is a rectangle with sides [tex]BC=\sqrt{3}[/tex] and [tex]CC_1=AA_1=\sqrt{AC^2+A_1C^2}=2\sqrt{3}[/tex] (right triangle [tex]ACA_1[/tex]), and [tex]S_{BCC_1B_1}=\sqrt{3}\times 2\sqrt{3}=6[/tex].
The total surface area of the prism is
[tex]S=2\times \frac{3\sqrt{2}}{2}+6\sqrt{2}+6+6=12+9\sqrt{2}[/tex]

Solution:
Let [tex]AD=BC=x[/tex], [tex]AB=CD=y[/tex] (so [tex]y\gt x[/tex]) and [tex]DM=h[/tex]. Since [tex]DM\perp(ABCD)[/tex], the triangles [tex]ADM[/tex], [tex]CDM[/tex] and [tex]BDM[/tex] are right-angled at [tex]D[/tex], and by the Pythagorean theorem
[tex]MA^2=h^2+x^2, \quad MC^2=h^2+y^2, \quad MB^2=h^2+BD^2=h^2+x^2+y^2[/tex]
Hence [tex]MB[/tex] is the longest of the three edges, and [tex]MC\gt MA[/tex] because [tex]y\gt x[/tex]. Therefore [tex]MB=5\sqrt{10}[/tex], [tex]MC=15[/tex], [tex]MA=13[/tex], and
[tex]x^2=MB^2-MC^2=250-225=25\Rightarrow x=5[/tex]
[tex]y^2=MB^2-MA^2=250-169=81\Rightarrow y=9[/tex]
[tex]h^2=MA^2-x^2=169-25=144\Rightarrow h=12[/tex]
The orthogonal projection of [tex]MA[/tex] on the base is [tex]DA[/tex] and [tex]AB\perp DA[/tex], so by the theorem of three perpendiculars [tex]AB\perp MA[/tex]. In the same way [tex]BC\perp DC[/tex] gives [tex]BC\perp MC[/tex]. Thus all four lateral faces are right triangles:
[tex]S_{ADM}=\frac{5\times 12}{2}=30, \quad S_{CDM}=\frac{9\times 12}{2}=54[/tex]
[tex]S_{ABM}=\frac{AB\times MA}{2}=\frac{9\times 13}{2}=\frac{117}{2}, \quad S_{BCM}=\frac{BC\times MC}{2}=\frac{5\times 15}{2}=\frac{75}{2}[/tex]
The base area is [tex]5\times 9=45[/tex], so the total surface area is
[tex]S=45+30+54+\frac{117}{2}+\frac{75}{2}=225[/tex]

Solution:
Let [tex]MO[/tex] be the height of the pyramid, [tex]O\in(ABCD)[/tex]. The right triangles [tex]AOM[/tex], [tex]BOM[/tex], [tex]COM[/tex], [tex]DOM[/tex] have the common leg [tex]MO[/tex] and equal hypotenuses [tex]MA=MB=MC=MD=65[/tex], so they are congruent and [tex]OA=OB=OC=OD[/tex]. Hence [tex]O[/tex] is the center of the circle circumscribed about the trapezoid [tex]ABCD[/tex].
A trapezoid inscribed in a circle is isosceles: [tex]\angle A+\angle D=180^{\circ}[/tex] because [tex]AB\parallel CD[/tex], and [tex]\angle A+\angle C=180^{\circ}[/tex] as opposite angles of a cyclic quadrilateral, so [tex]\angle C=\angle D[/tex] and [tex]AD=BC=30[/tex].
Let [tex]CH\perp AB[/tex], [tex]H\in AB[/tex]. Then [tex]BH=\frac{AB-CD}{2}=18[/tex], and from the right triangle [tex]BHC[/tex]
[tex]CH=\sqrt{BC^2-BH^2}=\sqrt{30^2-18^2}=24[/tex]
Let [tex]P[/tex] be the midpoint of [tex]AB[/tex]; then [tex]PA=PB=25[/tex] and [tex]PH=PB-BH=7[/tex], so from the right triangle [tex]PHC[/tex]
[tex]PC=\sqrt{PH^2+CH^2}=\sqrt{7^2+24^2}=25[/tex]
By the symmetry of the isosceles trapezoid, [tex]PD=PC=25[/tex]. Thus [tex]P[/tex] is equidistant from all four vertices, i.e. [tex]O=P[/tex] is the midpoint of [tex]AB[/tex] and [tex]OA=25[/tex]. From the right triangle [tex]AOM[/tex]
[tex]MO=\sqrt{MA^2-OA^2}=\sqrt{65^2-25^2}=\sqrt{3600}=60[/tex]

Solution:
Let [tex]AB=a[/tex] and [tex]BC=b[/tex]; then [tex]2(a+b)=22[/tex], i.e. [tex]a+b=11[/tex], and [tex]AA_1=CC_1=4[/tex].
Since [tex]AA_1\perp(ABC)[/tex], the projection of [tex]A_1B[/tex] on the base is [tex]AB[/tex], so the angle between [tex]A_1B[/tex] and the base is [tex]\varphi=\angle A_1BA[/tex]. In the same way the angle between [tex]C_1B[/tex] and the base is [tex]\psi=\angle C_1BC[/tex]. From the right triangles [tex]ABA_1[/tex] and [tex]CBC_1[/tex]
[tex]\tan \varphi=\frac{4}{a}, \quad \tan \psi=\frac{4}{b}[/tex]
The diagonals have different lengths, so [tex]a\neq b[/tex]; let [tex]a\gt b[/tex] (the other case is symmetric). Then [tex]\tan \varphi\lt\tan \psi[/tex], i.e. [tex]\varphi\lt\psi[/tex], hence [tex]\psi=2\varphi[/tex], where [tex]\varphi\lt 45^{\circ}[/tex] because [tex]\psi\lt 90^{\circ}[/tex]. Then
[tex]\frac{4}{b}=\tan 2\varphi=\frac{2\tan \varphi}{1-\tan ^2\varphi}=\frac{\frac{8}{a}}{1-\frac{16}{a^2}}=\frac{8a}{a^2-16}\Rightarrow a^2-16=2ab[/tex]
Substituting [tex]b=11-a[/tex], we get
[tex]a^2-16=22a-2a^2\iff 3a^2-22a-16=0\iff (a-8)(3a+2)=0[/tex]
Hence [tex]a=8[/tex] (the other root is negative) and [tex]b=3[/tex]. Check: [tex]\tan \varphi=\frac{1}{2}[/tex] and [tex]\tan 2\varphi=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}=\tan \psi[/tex]. The volume is
[tex]V=a\times b\times AA_1=8\times 3\times 4=96[/tex]

Solution:
Let the edges of the upper base and the lateral edges have length [tex]x[/tex], and the edges of the lower base length [tex]y[/tex]. Each lateral face, e.g. [tex]ABB_1A_1[/tex], is an isosceles trapezoid with bases [tex]y[/tex] and [tex]x[/tex], legs [tex]x[/tex] and height equal to the slant height [tex]4[/tex]. From the perimeter:
[tex]3x+y=26\Rightarrow y=26-3x[/tex]
Let [tex]A_1K[/tex] be the height of the trapezoid ([tex]K[/tex] on [tex]AB[/tex]). In an isosceles trapezoid [tex]AK=\frac{y-x}{2}=13-2x[/tex], and the Pythagorean theorem in the right triangle [tex]AKA_1[/tex] gives
[tex]x^2=4^2+(13-2x)^2\Rightarrow 3x^2-52x+185=0\Rightarrow x=5\text{ or }x=\frac{37}{3}[/tex]
The second root gives [tex]y=26-37\lt 0[/tex], so [tex]x=5[/tex] and [tex]y=11[/tex]. The lateral surface area and the areas of the bases are
[tex]S_L=4\times \frac{11+5}{2}\times 4=128,\quad S_{ABCD}=11^2=121,\quad S_{A_1B_1C_1D_1}=5^2=25[/tex]
Therefore the total surface area is
[tex]S=128+121+25=274[/tex]

Solution:
Let [tex]O[/tex] and [tex]O_1[/tex] be the centers of the bases [tex]ABC[/tex] and [tex]A_1B_1C_1[/tex]. Since the truncated pyramid is regular, [tex]OO_1=h[/tex] is its height. The circumradii of the equilateral bases are
[tex]OA=\frac{13\sqrt{3}}{3},\quad O_1A_1=\frac{\sqrt{3}}{3}[/tex]
The quadrilateral [tex]AOO_1A_1[/tex] is a right trapezoid. Drop the perpendicular [tex]A_1P[/tex] to [tex]AO[/tex]; then [tex]A_1P=h[/tex] and [tex]AP=OA-O_1A_1=4\sqrt{3}[/tex]. From the right triangle [tex]APA_1[/tex]:
[tex]h^2=AA_1^2-AP^2=64-48=16\Rightarrow h=4[/tex]
The base areas are [tex]S_1=\frac{13^2\sqrt{3}}{4}=\frac{169\sqrt{3}}{4}[/tex] and [tex]S_2=\frac{\sqrt{3}}{4}[/tex], so [tex]\sqrt{S_1S_2}=\frac{13\sqrt{3}}{4}[/tex]. By the volume formula for a truncated pyramid:
[tex]V=\frac{h}{3}\left(S_1+S_2+\sqrt{S_1S_2}\right)=\frac{4}{3}\times \frac{(169+1+13)\sqrt{3}}{4}=61\sqrt{3}[/tex]

Solution:
Let the perpendicular faces be [tex]ABB_1A_1[/tex] (area [tex]24[/tex]) and [tex]ACC_1A_1[/tex] (area [tex]30[/tex]); their common edge is [tex]AA_1[/tex]. A plane perpendicular to the lateral edges cuts the lines [tex]AA_1[/tex], [tex]BB_1[/tex], [tex]CC_1[/tex] at [tex]K[/tex], [tex]L[/tex], [tex]N[/tex]; the triangle [tex]KLN[/tex] is the perpendicular section of the prism.
The segment [tex]KL[/tex] lies in the face [tex]ABB_1A_1[/tex] and [tex]KL\perp AA_1[/tex], so it is the height of this parallelogram to the side [tex]AA_1=6[/tex]. Similarly for [tex]KN[/tex]:
[tex]KL=\frac{24}{6}=4,\quad KN=\frac{30}{6}=5[/tex]
Since [tex]KL\perp AA_1[/tex] and [tex]KN\perp AA_1[/tex], the angle [tex]LKN[/tex] is the linear angle of the dihedral angle between the two faces, so [tex]\angle LKN=90^{\circ}[/tex] and
[tex]S_{KLN}=\frac{1}{2}\times 4\times 5=10[/tex]
The volume of an oblique prism equals the area of its perpendicular section times the lateral edge (the part between the base [tex]ABC[/tex] and the section [tex]KLN[/tex] can be moved along the edges by [tex]AA_1[/tex] to the other base, which turns the prism into a right prism with base [tex]KLN[/tex] and height [tex]6[/tex]). Hence
[tex]V=S_{KLN}\times AA_1=10\times 6=60[/tex]

Solution:
Let [tex]AB=a[/tex], [tex]A_1B_1=b[/tex] ([tex]a\gt b[/tex]), let [tex]O[/tex] and [tex]O_1[/tex] be the centers of the bases ([tex]OO_1[/tex] is the height), and let [tex]K[/tex] and [tex]K_1[/tex] be the midpoints of [tex]AB[/tex] and [tex]A_1B_1[/tex]. The face [tex]ABB_1A_1[/tex] is an isosceles trapezoid, so [tex]KK_1\perp AB[/tex] and [tex]KK_1=k[/tex] is the slant height; also [tex]OK\perp AB[/tex]. Hence [tex]\angle K_1KO=\alpha[/tex] is the linear angle of the dihedral angle at [tex]AB[/tex].
In the right trapezoid [tex]OO_1K_1K[/tex] we have [tex]OK=\frac{a}{2}[/tex], [tex]O_1K_1=\frac{b}{2}[/tex]. Drop [tex]K_1H\perp OK[/tex]; from the right triangle [tex]KHK_1[/tex]:
[tex]KH=\frac{a-b}{2}=k\cos\alpha\Rightarrow k=\frac{a-b}{2\cos\alpha}[/tex]
The lateral surface consists of four congruent trapezoids:
[tex]S=4\times \frac{a+b}{2}\times k=\frac{a^2-b^2}{\cos\alpha}[/tex]
The condition [tex]S=a^2+b^2[/tex] gives
[tex]a^2-b^2=(a^2+b^2)\cos\alpha\Rightarrow a^2(1-\cos\alpha)=b^2(1+\cos\alpha)[/tex]
[tex]\frac{a^2}{b^2}=\frac{1+\cos\alpha}{1-\cos\alpha}=\frac{2\cos^2\frac{\alpha}{2}}{2\sin ^2\frac{\alpha}{2}}=\cot ^2\frac{\alpha}{2}[/tex]
Since [tex]\frac{\alpha}{2}[/tex] is acute, [tex]\frac{a}{b}=\cot \frac{\alpha}{2}[/tex], and the ratio of the perimeters is
[tex]\frac{4a}{4b}=\cot \frac{\alpha}{2}[/tex]

Solution:
Let [tex]DA=x[/tex], [tex]DB=y[/tex], [tex]DC=z[/tex]. The faces [tex]ABD[/tex], [tex]BCD[/tex] and [tex]CAD[/tex] are right triangles with the right angle at [tex]D[/tex], so
[tex]B_1=\frac{xy}{2},\quad B_2=\frac{yz}{2},\quad B_3=\frac{zx}{2}[/tex]
Multiplying these equalities, we get
[tex]B_1B_2B_3=\frac{x^2y^2z^2}{8}\Rightarrow xyz=\sqrt{8B_1B_2B_3}=2\sqrt{2B_1B_2B_3}[/tex]
The edge [tex]DA[/tex] is perpendicular to the two intersecting lines [tex]DB[/tex] and [tex]DC[/tex] of the plane [tex](BCD)[/tex], so [tex]DA\perp(BCD)[/tex] and [tex]DA=x[/tex] is the height of the pyramid to the face [tex]BCD[/tex], whose area is [tex]B_2=\frac{yz}{2}[/tex]. Therefore
[tex]V=\frac{1}{3}\times \frac{yz}{2}\times x=\frac{xyz}{6}=\frac{2\sqrt{2B_1B_2B_3}}{6}=\frac{1}{3}\sqrt{2B_1B_2B_3}[/tex]

Solution:
Let [tex]O[/tex] be the intersection point of the diagonals [tex]AC[/tex] and [tex]BD[/tex]; it is the midpoint of both. Triangle [tex]ABD[/tex] has [tex]AB=AD=a[/tex] and [tex]\angle BAD=60^{\circ}[/tex], so it is equilateral, [tex]BD=a[/tex] and its median is [tex]AO=\frac{a\sqrt{3}}{2}[/tex]; hence [tex]AC=2AO=a\sqrt{3}[/tex].
The point [tex]S[/tex] is not in the base plane, so [tex]SO[/tex] is a median of the triangles [tex]SAC[/tex] and [tex]SBD[/tex]. By the formula for the length of a median,
[tex]4SO^2=2SA^2+2SC^2-AC^2=2SB^2+2SD^2-BD^2[/tex]
Substituting [tex]SA^2=a^2[/tex], [tex]AC^2=3a^2[/tex] and [tex]BD^2=a^2[/tex]:
[tex]2a^2+2SC^2-3a^2=2SB^2+2SD^2-a^2\iff SC^2=SB^2+SD^2[/tex]
Hence [tex]SC\gt SB[/tex], [tex]SC\gt SD[/tex] and [tex]SC^2=SB^2+SD^2\lt(SB+SD)^2[/tex], i.e. [tex]SC\lt SB+SD[/tex], so the three segments satisfy the triangle inequalities. By the converse of the Pythagorean theorem, the triangle with sides [tex]SB[/tex], [tex]SD[/tex] and [tex]SC[/tex] is right-angled, with hypotenuse [tex]SC[/tex].

Solution:
Let [tex]ABCD[/tex] be the regular tetrahedron with height [tex]H[/tex], let [tex]P[/tex] be an interior point and [tex]d_1,d_2,d_3,d_4[/tex] its distances to the faces [tex]BCD[/tex], [tex]ACD[/tex], [tex]ABD[/tex], [tex]ABC[/tex]. All faces are congruent equilateral triangles, so they have the same area [tex]S[/tex].
The segments from [tex]P[/tex] to the vertices split the tetrahedron into the pyramids [tex]PBCD[/tex], [tex]PACD[/tex], [tex]PABD[/tex], [tex]PABC[/tex]; since [tex]P[/tex] is an interior point, they fill the tetrahedron without overlapping. Their bases are the faces of the tetrahedron and their heights are [tex]d_1,d_2,d_3,d_4[/tex], so
[tex]V_{ABCD}=\frac{1}{3}Sd_1+\frac{1}{3}Sd_2+\frac{1}{3}Sd_3+\frac{1}{3}Sd_4=\frac{1}{3}S\,(d_1+d_2+d_3+d_4)[/tex]
On the other hand, taking the base [tex]ABC[/tex] and the height [tex]H[/tex],
[tex]V_{ABCD}=\frac{1}{3}SH[/tex]
Comparing the two expressions and dividing by [tex]\frac{1}{3}S\neq 0[/tex], we get [tex]d_1+d_2+d_3+d_4=H[/tex] for every interior point [tex]P[/tex].

Solution:
Let the pyramid be [tex]ABCDM[/tex] with apex [tex]M[/tex], square base [tex]ABCD[/tex] of side [tex]a[/tex] and center [tex]O[/tex]. Since the pyramid is regular, the foot of its height is the center of the base: [tex]MO\perp(ABCD)[/tex].
Let [tex]K[/tex] be the midpoint of [tex]BC[/tex] and [tex]\gamma=\angle BMC[/tex]. The triangle [tex]BCM[/tex] is isosceles ([tex]MB=MC[/tex]), so its median [tex]MK[/tex] is also a height and a bisector: [tex]MK\perp BC[/tex] and [tex]\angle BMK=\frac{\gamma}{2}[/tex]. Also [tex]OK\parallel AB[/tex], hence [tex]OK\perp BC[/tex] and [tex]OK=\frac{a}{2}[/tex].
Therefore [tex]\angle MKO[/tex] is the linear angle of the dihedral angle at the edge [tex]BC[/tex], i.e. [tex]\angle MKO=\alpha[/tex]. From the right triangle [tex]MOK[/tex] (right angle at [tex]O[/tex]):
[tex]\cos\alpha=\frac{OK}{MK}=\frac{a}{2MK}[/tex]
From the right triangle [tex]BKM[/tex] (right angle at [tex]K[/tex], [tex]BK=\frac{a}{2}[/tex]):
[tex]\tan \frac{\gamma}{2}=\frac{BK}{MK}=\frac{a}{2MK}[/tex]
The two expressions are equal, hence [tex]\cos\alpha=\tan \frac{\gamma}{2}[/tex].