Prisms and Pyramids Problems: Problems with Solutions

Solution:
Let the parallelepiped be [tex]ABCDA_1B_1C_1D_1[/tex] with rhombus base [tex]ABCD[/tex], [tex]AC=8[/tex], [tex]BD=6[/tex], and let [tex]O[/tex] be the intersection point of the diagonals of the base. The diagonals of a rhombus are perpendicular and bisect each other, so triangle [tex]ABO[/tex] is right-angled at [tex]O[/tex] with [tex]AO=4[/tex], [tex]BO=3[/tex]:
[tex]AB=\sqrt{AO^2+BO^2}=\sqrt{4^2+3^2}=5[/tex]
The parallelepiped is right, so its lateral faces are congruent rectangles with sides [tex]5[/tex] and the height; let [tex]AB_1=13[/tex]. From the right triangle [tex]ABB_1[/tex] (right angle at [tex]B[/tex]) we get the height:
[tex]BB_1=\sqrt{AB_1^2-AB^2}=\sqrt{13^2-5^2}=12[/tex]
The base area is half the product of the diagonals, and the lateral surface area is the perimeter of the base times the height. Therefore the total surface area is
[tex]S_{ABCD}=\frac{AC\times BD}{2}=\frac{8\times 6}{2}=24,\quad S=2S_{ABCD}+4\times AB\times BB_1=48+4\times 5\times 12=48+240=288[/tex]

Solution:
Let the dimensions be [tex]n-1[/tex], [tex]n[/tex] and [tex]n+1[/tex], where [tex]n\ge 2[/tex] is an integer. Then
[tex]V=(n-1)n(n+1)=n^3-n=24\iff n^3-n-24=0\iff (n-3)(n^2+3n+8)=0[/tex]
The discriminant of [tex]n^2+3n+8[/tex] is [tex]9-32\lt 0[/tex], so [tex]n^2+3n+8\gt 0[/tex] and the only solution is [tex]n=3[/tex]. Hence the dimensions are [tex]a=2[/tex], [tex]b=3[/tex], [tex]c=4[/tex] (check: [tex]2\times 3\times 4=24[/tex]), and the total surface area is
[tex]S=2(ab+bc+ca)=2(2\times 3+3\times 4+4\times 2)=2\times 26=52[/tex]

Solution:
The volume of the metal does not change, so the volume of the new cube is the sum of the three volumes:
[tex]V=3^3+4^3+5^3=27+64+125=216[/tex]
If [tex]a[/tex] is the edge of the new cube, then [tex]a^3=216=6^3[/tex], so [tex]a=6[/tex]. The surface of a cube consists of six squares with side [tex]a[/tex]:
[tex]S=6a^2=6\times 36=216[/tex]

Solution:
Let [tex]ABCDA_1B_1C_1D_1[/tex] be the truncated pyramid, whose bases are the squares [tex]ABCD[/tex] ([tex]AB=16[/tex]) and [tex]A_1B_1C_1D_1[/tex] ([tex]A_1B_1=4[/tex]) with centers [tex]O[/tex] and [tex]O_1[/tex]; the height is [tex]OO_1=h[/tex]. Let [tex]K[/tex] and [tex]K_1[/tex] be the midpoints of [tex]BC[/tex] and [tex]B_1C_1[/tex]. The lateral face [tex]BCC_1B_1[/tex] is an isosceles trapezoid, so its height [tex]KK_1=10[/tex] is the slant height.
The points [tex]O[/tex], [tex]K[/tex], [tex]K_1[/tex], [tex]O_1[/tex] lie in one plane (through [tex]OO_1[/tex], perpendicular to [tex]BC[/tex]), and [tex]OKK_1O_1[/tex] is a right trapezoid with [tex]OK=\frac{AB}{2}=8[/tex] and [tex]O_1K_1=\frac{A_1B_1}{2}=2[/tex]. Drop [tex]K_1P\perp OK[/tex]; then [tex]K_1P=h[/tex] and [tex]PK=8-2=6[/tex]. From the right triangle [tex]K_1PK[/tex]:
[tex]h=\sqrt{KK_1^2-PK^2}=\sqrt{10^2-6^2}=8[/tex]
The areas of the bases are [tex]S_1=16^2=256[/tex] and [tex]S_2=4^2=16[/tex], so by the formula for the volume of a truncated pyramid
[tex]V=\frac{h}{3}\left(S_1+S_2+\sqrt{S_1S_2}\right)=\frac{8}{3}(256+16+64)=\frac{8}{3}\times 336=896[/tex]

Solution:
Let [tex]ABCDA_1B_1C_1D_1[/tex] be the prism, [tex]AB=6[/tex], [tex]AD=3[/tex], [tex]\angle DAB=45^{\circ}[/tex]. The area of the base is
[tex]S=AB\times AD\times \sin 45^{\circ}=6\times 3\times \frac{\sqrt{2}}{2}=9\sqrt{2}[/tex]
Let [tex]H[/tex] be the orthogonal projection of [tex]A_1[/tex] on the plane of the base. Then [tex]A_1H=h[/tex] is the height of the prism, [tex]AH[/tex] is the projection of the lateral edge [tex]AA_1[/tex], and the angle between the lateral edge and the base is [tex]\angle A_1AH=30^{\circ}[/tex]. In the right triangle [tex]AHA_1[/tex] the leg [tex]A_1H[/tex] lies opposite the [tex]30^{\circ}[/tex] angle, so
[tex]h=A_1H=\frac{AA_1}{2}=2\sqrt{2}[/tex]
The volume is the base area times the height (not times the lateral edge):
[tex]V=S\times h=9\sqrt{2}\times 2\sqrt{2}=36[/tex]

Solution:
Let [tex]ABCA_1B_1C_1[/tex] be the prism and [tex]AA_1=BB_1=CC_1=l[/tex]. Each lateral face is a parallelogram; for example, in [tex]ABB_1A_1[/tex] the sides [tex]AA_1[/tex] and [tex]BB_1[/tex] have length [tex]l[/tex], and the height to them is the distance between the parallel lines [tex]AA_1[/tex] and [tex]BB_1[/tex].
Hence the areas of the three lateral faces are [tex]2l[/tex], [tex]3l[/tex] and [tex]4l[/tex], and
[tex]2l+3l+4l=45\Rightarrow 9l=45\Rightarrow l=5[/tex]

Solution:
The total surface area [tex]S_1[/tex] of a prism is the sum of its lateral surface area [tex]S[/tex] and the areas of its two bases:
[tex]S_1=S+2S_{ABCD}[/tex]
The base is a trapezoid with bases [tex]AB[/tex], [tex]CD[/tex] and height [tex]5[/tex], so
[tex]S_{ABCD}=\frac{AB+CD}{2}\times 5=\frac{14+6}{2}\times 5=50[/tex]
Hence the lateral surface area is
[tex]S=S_1-2S_{ABCD}=450-2\times 50=350[/tex]
Note that neither the legs of the trapezoid nor the height of the prism are needed.

Solution:
Let [tex]O[/tex] be the foot of the height from [tex]D[/tex] to the plane [tex]ABC[/tex]. The right triangles [tex]DOA[/tex], [tex]DOB[/tex] and [tex]DOC[/tex] have the common leg [tex]DO[/tex] and equal hypotenuses [tex]DA=DB=DC[/tex], so they are congruent and [tex]OA=OB=OC[/tex].
Hence [tex]O[/tex] is the circumcenter of [tex]\triangle ABC[/tex] and [tex]OA=R[/tex] is its circumradius. By the law of sines in [tex]\triangle ABC[/tex]:
[tex]AB=2R\sin \angle ACB\Rightarrow 3=2R\times \frac{1}{2}\Rightarrow R=3[/tex]
From the right triangle [tex]DOA[/tex]:
[tex]DO=\sqrt{DA^2-OA^2}=\sqrt{25-9}=4[/tex]
Note that the triangle [tex]ABC[/tex] itself is not determined by the data, but the height is.

Solution:
Let [tex]O[/tex] be the intersection point of the diagonals of the square [tex]ABCD[/tex]. The pyramid is regular, so its height is [tex]MO[/tex] and [tex]AO[/tex] is the orthogonal projection of [tex]AM[/tex] onto the base; hence the angle between [tex]AM[/tex] and the base is [tex]\angle MAO=45^{\circ}[/tex].
The right triangle [tex]AOM[/tex] (right angle at [tex]O[/tex]) has an acute angle of [tex]45^{\circ}[/tex], so it is isosceles:
[tex]MO=AO=\frac{AM}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}}=3[/tex]
Then the diagonal of the square is [tex]AC=2AO=6[/tex], and the area of the base is
[tex]B=\frac{AC^2}{2}=\frac{36}{2}=18[/tex]
Therefore
[tex]V=\frac{1}{3}B\times MO=\frac{1}{3}\times 18\times 3=18[/tex]

Solution:
The lateral faces are squares, so every lateral edge is equal to the base edge; denote this common length by [tex]a[/tex].
A hexagonal prism has [tex]6+6=12[/tex] base edges (on the two bases) and [tex]6[/tex] lateral edges, i.e. [tex]18[/tex] edges in total, all of length [tex]a[/tex]. Hence
[tex]18a=90\Rightarrow a=5[/tex]
The lateral surface consists of [tex]6[/tex] equal squares with side [tex]a=5[/tex], so
[tex]S=6a^2=6\times 5^2=150[/tex]

Solution:
Let the pyramid be [tex]ABCDM[/tex] with apex [tex]M[/tex], base center [tex]O[/tex] and base edge [tex]a[/tex]. Let [tex]K[/tex] be the midpoint of [tex]BC[/tex]; then [tex]MK=k[/tex] is the apothem (the height of the isosceles triangle [tex]BCM[/tex]) and [tex]MO=h=\frac{k}{2}[/tex] is the height of the pyramid.
Since [tex]MO[/tex] is perpendicular to the base, the triangle [tex]MOK[/tex] is right-angled at [tex]O[/tex], with [tex]OK=\frac{a}{2}[/tex]. By the Pythagorean theorem
[tex]k^2=h^2+\frac{a^2}{4}=\frac{k^2}{4}+\frac{a^2}{4}\Rightarrow k=\frac{a}{\sqrt{3}}, \ h=\frac{a}{2\sqrt{3}}[/tex]
The lateral surface consists of four congruent isosceles triangles with base [tex]a[/tex] and height [tex]k[/tex], and the volume is one third of the base area times the height:
[tex]S=4\times \frac{1}{2}ak=\frac{2a^2}{\sqrt{3}}, \quad V=\frac{1}{3}a^2h=\frac{a^3}{6\sqrt{3}}[/tex]
The condition [tex]S=6V[/tex] gives
[tex]\frac{2a^2}{\sqrt{3}}=\frac{a^3}{\sqrt{3}}\Rightarrow a=2[/tex]

Solution:
Let the base edge be [tex]a[/tex] and the height be [tex]h[/tex]. The base [tex]ABCD[/tex] is a square and the lateral edges are perpendicular to the base. From the right triangle [tex]ABB_1[/tex] (right angle at [tex]B[/tex]):
[tex]a^2+h^2=AB_1^2=25[/tex]
The diagonal of the square is [tex]AC=a\sqrt{2}[/tex], and from the right triangle [tex]ACC_1[/tex] (right angle at [tex]C[/tex]):
[tex]2a^2+h^2=AC_1^2=49[/tex]
Subtracting the first equation from the second gives [tex]a^2=24[/tex], hence [tex]h^2=1[/tex] and [tex]h=1[/tex]. The volume is
[tex]V=a^2h=24\times 1=24[/tex]

Solution:
Let the pyramid be [tex]ABCM[/tex] with base the equilateral triangle [tex]ABC[/tex] and height [tex]MO[/tex]. Since the pyramid is regular, [tex]O[/tex] is the center of [tex]\triangle ABC[/tex], i.e. the intersection point of its medians.
Let [tex]K[/tex] be the midpoint of [tex]AB[/tex]. Then [tex]O[/tex] lies on the median [tex]CK[/tex], and since the centroid divides each median in ratio [tex]2:1[/tex] from the vertex,
[tex]OC=2OK[/tex]
The orthogonal projection of the lateral edge [tex]MC[/tex] on the base is [tex]OC[/tex], so [tex]\angle MCO=\beta[/tex], and from the right triangle [tex]MOC[/tex]:
[tex]\tan \beta=\frac{MO}{OC}[/tex]
The face [tex]ABM[/tex] is an isosceles triangle, so its median [tex]MK[/tex] is also a height: [tex]MK\perp AB[/tex]. Also [tex]OK\perp AB[/tex], since [tex]CK[/tex] is a height of the equilateral triangle [tex]ABC[/tex]. Hence [tex]\angle MKO[/tex] is the linear angle of the dihedral angle at the edge [tex]AB[/tex], i.e. [tex]\angle MKO=\alpha[/tex], and from the right triangle [tex]MOK[/tex]:
[tex]\tan \alpha=\frac{MO}{OK}[/tex]
Therefore
[tex]\tan \alpha=\frac{MO}{OK}=\frac{2MO}{OC}=2\tan \beta[/tex]

Solution:
Let [tex]O[/tex] be the orthogonal projection of [tex]D[/tex] onto the plane [tex](ABC)[/tex]. A line perpendicular to a plane is perpendicular to every line of the plane, so [tex]DO\perp OA[/tex], [tex]DO\perp OB[/tex], [tex]DO\perp OC[/tex]. By the Pythagorean theorem in the right triangles [tex]DOA[/tex], [tex]DOB[/tex], [tex]DOC[/tex]:
[tex]OA^2=DA^2-DO^2,\quad OB^2=DB^2-DO^2,\quad OC^2=DC^2-DO^2[/tex]
Since [tex]DA=DB=DC[/tex], it follows that
[tex]OA=OB=OC[/tex]
The point [tex]O[/tex] cannot coincide with a vertex, e.g. with [tex]A[/tex], because then [tex]OB=OA=0[/tex], i.e. [tex]B=A[/tex]. So [tex]O[/tex] is a point of the plane [tex](ABC)[/tex] equidistant from [tex]A[/tex], [tex]B[/tex] and [tex]C[/tex], i.e. the center of the circle circumscribed about triangle [tex]ABC[/tex].