Diketahui kubus ABCD.EFGH dengan panjang rusuk AB=9 cm.

Irisan kubus dengan bidang PQR membentuk sebuah heksagon (segi enam) beraturan, dengan luas $\frac{243\sqrt{3}}{4}$ cm$^2$.

Pembahasan

a. Untuk melukis irisan kubus dengan bidang PQR, kita perlu menemukan perpotongan bidang PQR dengan sisi-sisi kubus.
P adalah titik tengah AD, Q adalah titik tengah AB, dan R adalah titik tengah BF.
Bidang PQR akan memotong rusuk AB di Q, rusuk AD di P.
Kita perlu mencari titik potong lain. Perhatikan bidang ABFE. Garis PQ terletak pada bidang ADHE dan bidang ABFE. Perpanjang PQ hingga memotong perpanjangan EH di S. Maka S adalah titik potong garis PQ dengan bidang EFGH.
Perhatikan bidang BCGF. Titik R ada pada BF. Bidang PQR akan memotong BCGF sepanjang garis yang melalui R dan sejajar PQ. Karena PQ sejajar AF, maka garis potongnya sejajar AF.
Perhatikan bidang CDHG. Titik P ada pada AD. Bidang PQR akan memotong CDHG pada garis yang melalui P dan sejajar QR.
Cara lain: Cari titik potong bidang PQR dengan rusuk-rusuk kubus.
Titik Q pada AB, P pada AD.
Dari P, tarik garis sejajar QR, memotong HG di S.
Dari Q, tarik garis sejajar PR, memotong AE di T.
Irisan PQRS adalah jajar genjang.
b. Bentuk penampang adalah jajar genjang. Bukti: PQ sejajar dengan garis yang menghubungkan titik tengah AE dan AD, yaitu sejajar dengan diagonal AH.
QR sejajar dengan garis yang menghubungkan titik tengah AB dan BF, yaitu sejajar dengan AF.
PS sejajar dengan QR, maka PS sejajar AF.
PR sejajar dengan diagonal BH.
PT sejajar dengan diagonal BG.
Karena P di AD, Q di AB, R di BF, maka PQRS akan membentuk jajar genjang jika PS sejajar QR dan PQ sejajar SR.
Kita tahu PQ sejajar FH.
Jika kita mengambil titik potong S pada EH, maka PQRS akan menjadi jajar genjang.
Lebih tepatnya, mari kita tentukan titik-titik potongnya:
1. Bidang PQR memotong AB di Q.
2. Bidang PQR memotong AD di P.
3. Perpanjangan PQ memotong EH di S.
4. SR sejajar PQ, maka SR memotong FG di R' (titik tengah FG).
Irisannya adalah PQSR' (jajar genjang).
c. Menghitung luas bidang irisan PQSR'.
Misalkan panjang rusuk kubus adalah $a=9$ cm.
Q adalah titik tengah AB, maka $AQ = QB = \frac{a}{2} = 4.5$ cm.
P adalah titik tengah AD, maka $AP = PD = \frac{a}{2} = 4.5$ cm.
R adalah titik tengah BF, maka $BR = RF = \frac{a}{2} = 4.5$ cm.
Panjang PQ: Dalam segitiga siku-siku APQ, $PQ^2 = AP^2 + AQ^2 = (4.5)^2 + (4.5)^2 = 2 \times (4.5)^2$. $PQ = 4.5\sqrt{2}$ cm.
Panjang PS: Kita perlu mencari titik S pada EH. PQRS adalah jajar genjang, maka $PS = QR$. $QR^2 = QB^2 + BR^2 = (4.5)^2 + (4.5)^2 = 2 \times (4.5)^2$. $QR = 4.5\sqrt{2}$ cm.
Jadi, $PQ = QR = 4.5\sqrt{2}$ cm. PQRS adalah belah ketupat.
Untuk membuktikan ini, kita perlu melihat sudutnya. Sudut antara PQ dan PS.
$PQ$ arahnya $\frac{1}{2} \vec{AB} + \frac{1}{2} \vec{AD}$
$PS$ arahnya sejajar $QR$, $QR$ arahnya $\frac{1}{2} \vec{AB} + \frac{1}{2} \vec{BF}$.
Karena PQ sejajar FH dan QR sejajar BG, maka PQRS adalah jajar genjang.
Jika PQRS adalah belah ketupat, maka diagonalnya tegak lurus. Diagonalnya adalah PR dan QS.
$PR^2 = PQ^2 + QR^2 - 2 PQ QR \cos(\angle PQR)$.
Untuk menghitung luas, kita bisa menggunakan luas jajar genjang $Luas = alas \times tinggi$. Atau jika kita tahu diagonalnya, $Luas = \frac{1}{2} d_1 d_2$. Atau jika kita tahu dua sisi dan sudut apitnya $Luas = sisi1 \times sisi2 \sin(\text{sudut})$.
Kita tahu $PQ = 4.5\sqrt{2}$ dan $QR = 4.5\sqrt{2}$.
Mari kita cari sudut antara PQ dan QR.
$PQ$ adalah garis yang menghubungkan titik tengah AD dan AB.
$QR$ adalah garis yang menghubungkan titik tengah AB dan BF.
Dalam kubus, sudut antara rusuk-rusuknya adalah 90 derajat.
$PQ$ tegak lurus dengan $QR$? Tidak selalu.
$PQ$ sejajar dengan diagonal FH. $QR$ sejajar dengan diagonal BG.
Sudut antara FH dan BG tidak harus 90 derajat.
Mari kita cari luas menggunakan luas segitiga.
Luas PQRS = Luas $\triangle PQR$ + Luas $\triangle PSR$.
Karena PQRS jajar genjang, Luas PQRS = 2 x Luas $\triangle PQR$.
Kita bisa menggunakan rumus Heron jika kita tahu ketiga sisinya. Kita tahu PQ dan QR. Kita perlu PR.
$PR^2 = PB^2 + BR^2$? P bukan di AD, P di AD. Q di AB. R di BF.
$PB^2 = AB^2 + AP^2 = a^2 + (a/2)^2 = a^2 + a^2/4 = 5a^2/4$. $PB = \frac{a\sqrt{5}}{2}$.
$PR^2 = PB^2 + BR^2$? Bukan segitiga siku-siku.
Mari kita gunakan vektor.
A=(0,0,0), B=(a,0,0), D=(0,a,0), F=(a,0,a)
Q = (a/2, 0, 0)
P = (0, a/2, 0)
R = (a, 0, a/2)
$\\vec{QP} = P - Q = (-a/2, a/2, 0)$
$\\vec{QR} = R - Q = (a/2, 0, a/2)$
$|PQ| = \sqrt{(-a/2)^2 + (a/2)^2 + 0^2} = \sqrt{a^2/4 + a^2/4} = \sqrt{a^2/2} = \frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}$
$|QR| = \sqrt{(a/2)^2 + 0^2 + (a/2)^2} = \sqrt{a^2/4 + a^2/4} = \sqrt{a^2/2} = \frac{a\sqrt{2}}{2}$
Karena $|PQ| = |QR|$, maka PQRS adalah belah ketupat.
Luas belah ketupat = $\frac{1}{2} d_1 d_2$
$d_1 = PR = |R-P| = |(a, -a/2, a/2)| = \sqrt{a^2 + (-a/2)^2 + (a/2)^2} = \sqrt{a^2 + a^2/4 + a^2/4} = \sqrt{a^2 + a^2/2} = \sqrt{3a^2/2} = a\sqrt{3/2} = \frac{a\sqrt{6}}{2}$
$d_2 = QS = |S-Q|$. Kita perlu S. S adalah perpanjangan PQ memotong EH. Garis PQ: $Q + t(P-Q) = (a/2, 0, 0) + t(-a/2, a/2, 0) = (a/2 - ta/2, ta/2, 0)$.
Titik pada EH memiliki y=a, z=a. EH: $E + u(H-E)$. E=(0,a,0), H=(a,a,0). Wait, E=(0,a,0), F=(a,a,0), G=(a,a,a), H=(0,a,a). Let's use standard coordinates. A=(0,0,0), B=(a,0,0), C=(a,a,0), D=(0,a,0), E=(0,0,a), F=(a,0,a), G=(a,a,a), H=(0,a,a).
Q (titik tengah AB) = (a/2, 0, 0)
P (titik tengah AD) = (0, a/2, 0)
R (titik tengah BF) = (a, 0, a/2)
$\\vec{QP} = P - Q = (-a/2, a/2, 0)$
$\\vec{QR} = R - Q = (a, 0, a/2) - (a/2, 0, 0) = (a/2, 0, a/2)$
$|PQ| = \sqrt{(-a/2)^2 + (a/2)^2 + 0^2} = \sqrt{a^2/4 + a^2/4} = \sqrt{a^2/2} = \frac{a\sqrt{2}}{2}$
$|QR| = \sqrt{(a/2)^2 + 0^2 + (a/2)^2} = \sqrt{a^2/4 + a^2/4} = \sqrt{a^2/2} = \frac{a\sqrt{2}}{2}$
PQRS is a rhombus.
Diagonal 1: PR. $PR = |R-P| = |(a, 0, a/2) - (0, a/2, 0)| = |(a, -a/2, a/2)| = \sqrt{a^2 + (-a/2)^2 + (a/2)^2} = \sqrt{a^2 + a^2/4 + a^2/4} = \sqrt{a^2 + a^2/2} = \sqrt{3a^2/2} = \frac{a\sqrt{6}}{2}$
Diagonal 2: QS. Perlu cari titik S. Titik S terletak pada bidang PQR. Garis PQ: $Q + t \vec{QP} = (a/2, 0, 0) + t(-a/2, a/2, 0) = (a/2(1-t), a/2 t, 0)$.
Bidang PQR. Vektor normal bidang PQR? $\\vec{QP} \times \vec{QR} = (-a/2, a/2, 0) \times (a/2, 0, a/2) = (a^2/4, a^2/4, -a^2/4)$. Vektor normal $(1, 1, -1)$.
Persamaan bidang PQR: $1(x - a/2) + 1(y - 0) - 1(z - 0) = 0 
=> x - a/2 + y - z = 0 
=> x + y - z = a/2$.
Titik S terletak pada bidang EFGH (y=a, z=a) dan pada garis PQ yang diperpanjang.
Titik S pada EH. EH adalah garis $(0,a,0)$ ke $(0,a,a)$. S = $(0,a,z_s)$.
S juga terletak pada bidang PQR: $0 + a - z_s = a/2 => z_s = a/2$. Jadi S = (0, a, a/2).
QS = $|S-Q| = |(0, a, a/2) - (a/2, 0, 0)| = |(-a/2, a, a/2)| = \sqrt{(-a/2)^2 + a^2 + (a/2)^2} = \sqrt{a^2/4 + a^2 + a^2/4} = \sqrt{a^2/2 + a^2} = \sqrt{3a^2/2} = \frac{a\sqrt{6}}{2}$.
Oops, diagonalnya sama panjang. PQRS adalah persegi.
$|PQ| = \frac{a\sqrt{2}}{2}$, $|QR| = \frac{a\sqrt{2}}{2}$.
Sudut antara PQ dan QR? $\\vec{QP} \cdot \\vec{QR} = (-a/2, a/2, 0) \cdot (a/2, 0, a/2) = -a^2/4$. Cosinus sudutnya $\frac{-a^2/4}{|PQ||QR|} = \frac{-a^2/4}{(a\sqrt{2}/2)(a\sqrt{2}/2)} = \frac{-a^2/4}{a^2/2} = -1/2$. Sudutnya 120 derajat.
Ah, perpanjangan PQ memotong EH. Garis PQ melalui Q=(a/2,0,0) dan P=(0,a/2,0). Vektor arahnya $(-a/2, a/2, 0)$.
Garis PQ: $(a/2, 0, 0) + t(-a/2, a/2, 0) = (a/2(1-t), a/2 t, 0)$.
Titik pada EH memiliki koordinat $(x_e, a, z_e)$. Garis EH adalah $(0, a, 0)$ ke $(0, a, a)$. Jadi titik pada EH adalah $(0, a, z)$.
Kita cari t sehingga $(a/2(1-t), a/2 t, 0) = (0, a, z)$.
$a/2(1-t) = 0 => 1-t = 0 => t = 1$. Titik PQ pada t=1 adalah Q itu sendiri.
Sepertinya saya salah mendefinisikan titik.
Kubus ABCD.EFGH. AB rusuk bawah, EF rusuk atas. A=(0,0,0), B=(a,0,0), C=(a,a,0), D=(0,a,0), E=(0,0,a), F=(a,0,a), G=(a,a,a), H=(0,a,a).
P tengah AD = (0, a/2, 0)
Q tengah AB = (a/2, 0, 0)
R tengah BF = (a, 0, a/2)
$\\vec{QP} = P-Q = (-a/2, a/2, 0)$
$\\vec{QR} = R-Q = (a/2, 0, a/2)$
$|PQ| = \frac{a\sqrt{2}}{2}$
$|QR| = \frac{a\sqrt{2}}{2}$
PR tegak lurus QS?
$PR = |R-P| = |(a, -a/2, a/2)| = \frac{a\sqrt{6}}{2}$
QS = $|S-Q|$. S adalah perpotongan PQR dengan EH.
Titik S pada EH. EH adalah garis $(0,0,a)$ ke $(0,a,a)$? No. EH is $(0,0,a)$ to $(0,a,a)$? No. E=(0,0,a), H=(0,a,a). EH is the line $(0, y, a)$ for $0 \	ext{le } y \	ext{ le } a$. EH is along the y-axis in the plane x=0.
Let's re-orient. A=(0,0,0), B=(a,0,0), D=(0,a,0), E=(0,0,a). Then F=(a,0,a), H=(0,a,a), G=(a,a,a).
P tengah AD = (0, a/2, 0)
Q tengah AB = (a/2, 0, 0)
R tengah BF = (a, 0, a/2)
$\\vec{QP} = (-a/2, a/2, 0)$
$\\vec{QR} = (a/2, 0, a/2)$
$|PQ| = \frac{a\sqrt{2}}{2}$
$|QR| = \frac{a\sqrt{2}}{2}$
PR = $|R-P| = |(a, -a/2, a/2)| = \frac{a\sqrt{6}}{2}$
QS = $|S-Q|$. S lies on EH. EH is the line $(0, y, a)$ where $0 	ext{ le } y 	ext{ le } a$. So S = (0, y_s, a).
S must lie on the plane PQR. Plane equation: $n \cdot (x-x_0, y-y_0, z-z_0) = 0$.
$n = \vec{QP} \times \vec{QR} = (-a/2, a/2, 0) \times (a/2, 0, a/2) = (a^2/4, a^2/4, -a^2/4)$. Normal vector $(1, 1, -1)$.
Plane: $1(x - a/2) + 1(y - 0) - 1(z - 0) = 0 
=> x + y - z = a/2$.
Substitute S=(0, y_s, a): $0 + y_s - a = a/2 => y_s = 3a/2$. This point is outside the segment EH (where $y$ goes from 0 to $a$).
My coordinate system or assumption about S is wrong.
Let's use the geometry. P in AD, Q in AB, R in BF.
PQ connects midpoints of AD, AB. PQ is parallel to DB. Length $PQ = \frac{1}{2} DB = \frac{1}{2} a\sqrt{2} = \frac{a\sqrt{2}}{2}$.
QR connects midpoints of AB, BF. QR is parallel to AF. Length $QR = \frac{1}{2} AF = \frac{1}{2} a\sqrt{2} = \frac{a\sqrt{2}}{2}$.
So PQ = QR. PQRS is a rhombus.
We need the intersection of plane PQR with the rest of the cube edges.
Line PQ intersects EH at S. Since PQ is parallel to DB, and DB is parallel to FH, PQ is parallel to FH.
Consider the plane containing DBFH. This plane cuts the cube. Our plane PQR intersects this plane along the line passing through Q and parallel to FH. So PQ is parallel to FH.
Let's consider the face ABFE. P is on AD, Q is on AB, R is on BF. The plane cuts AB at Q and BF at R.
Consider the face ADHE. The plane cuts AD at P. Let it intersect EH at S.
Since PQRS is a rhombus, PS must be parallel to QR. QR is parallel to AF. So PS must be parallel to AF.
Since AF is perpendicular to AB and AE, PS must be perpendicular to AD and AE (if it were on the same plane).
Let's find S. Plane PQR intersects EH at S. Consider the projection onto the base ABCD. P is midpoint of AD, Q is midpoint of AB. The line PQ is drawn. Consider the projection onto the front face ABFE. Q is midpoint of AB, R is midpoint of BF. The line QR is drawn.
Let's think about symmetry. The plane passes through Q (midpoint of AB) and R (midpoint of BF). It also passes through P (midpoint of AD).
Consider the line through P parallel to QR. Since QR is parallel to AF, this line is parallel to AF. This line intersects the plane containing CDHG. Let's call this intersection point S'. Then PQRS' is a rhombus.
Let's use the first coordinate system where A=(0,0,0), B=(a,0,0), D=(0,a,0), E=(0,0,a).
P=(0, a/2, 0), Q=(a/2, 0, 0), R=(a, 0, a/2).
Line PQ: $(a/2, 0, 0) + t(-a/2, a/2, 0) = (a/2(1-t), a/2 t, 0)$.
EH: $E=(0,0,a), H=(0,a,a)$. Line EH is $(0, y, a)$ for $0 	ext{ le } y 	ext{ le } a$. So S = $(0, y_s, a)$.
For S to be on line PQ extended, we need $z=0$. But S has $z=a$. This means PQ does not intersect EH extended in a way that forms the shape.
Let's reconsider the description of the intersection. Plane PQR intersects the cube. It intersects face ABFE at QR. It intersects face ABCD at PQ. It intersects face BCGF at QR. It intersects face ADHE at P and some other point S on EH.
Consider the plane containing P, Q, R. P is on AD, Q is on AB, R is on BF.
Line PQ is in plane ABCD. Line QR is in plane BCGF.
Line PR connects midpoint of AD to midpoint of BF. This line is skew to AB.
Let's find the intersection with edge EH. Let S be the intersection point on EH.
Since P is on AD, Q is on AB, R is on BF, S is on EH.
Consider the face ADHE. P is the midpoint of AD. The plane cuts AD at P. Let it cut EH at S. The segment PS lies in the plane PQR and in the face ADHE.
Consider the face ABFE. The plane cuts AB at Q and BF at R. The segment QR lies in the plane PQR and in the face ABFE.
Consider the face BCGF. The plane cuts BF at R. Let it cut CG at T. The segment RT lies in the plane PQR and in the face BCGF.
Consider the face CDHG. Let it cut DH at U and HG at V. The segment PU lies in the plane PQR and in face ADHE. The segment PV lies in plane PQR and face CDHG.
This is getting complicated. Let's use the property that opposite sides of the intersection polygon are parallel.
PQ is in the base plane. Let's find a line parallel to PQ in the top plane EFGH. This line must pass through S on EH and intersect FG at some point.
Since PQ is parallel to DB, and DB is parallel to FH, PQ is parallel to FH. So the line through S on EH, parallel to PQ, must also be parallel to FH.
Let S be on EH. Then the line through S parallel to PQ must intersect FG. Let this point be S'.
So the intersection is PQRS' where S' is on FG.
Is S' the same as R? No. R is on BF.
Let's restart with the description of the shape. P on AD, Q on AB, R on BF. PQRS is the intersection.
$\\vec{PQ}$ is parallel to $\\vec{DB}$. $\\vec{QR}$ is parallel to $\\vec{AF}$.
Since $\\vec{DB}$ is not parallel to $\\vec{AF}$, PQRS is not a parallelogram unless some conditions are met.
Ah, the question asks to prove the shape. Let's assume the intersection is a polygon PQRS.
P on AD, Q on AB, R on BF. Let the plane PQR intersect edge EH at S and edge CG at T. Then PQRS is the intersection.
$PQ = \frac{a\sqrt{2}}{2}$, $QR = \frac{a\sqrt{2}}{2}$.
Also, $PS$ must be parallel to $QR$, so $PS$ parallel to $AF$. $PS = QR = \frac{a\sqrt{2}}{2}$.
And $RS$ must be parallel to $PQ$, so $RS$ parallel to $DB$. $RS = PQ = \frac{a\sqrt{2}}{2}$.
Thus PQRS is a rhombus.
b. The shape of the cross-section is a rhombus. Proof: P, Q, R are midpoints of AD, AB, BF respectively. $PQ = \frac{1}{2} DB = \frac{a\sqrt{2}}{2}$. $QR = \frac{1}{2} AF = \frac{a\sqrt{2}}{2}$. Let S be the intersection of plane PQR with EH. Since PQRS is a planar section, $PS$ must be parallel to $QR$ and $RS$ must be parallel to $PQ$. Thus, $PS = QR = \frac{a\sqrt{2}}{2}$ and $RS = PQ = \frac{a\sqrt{2}}{2}$. Since all four sides are equal, PQRS is a rhombus.
c. To calculate the area of the rhombus PQRS, we need the length of the diagonals or the angle between adjacent sides.
We have $|PQ| = \frac{a\sqrt{2}}{2}$ and $|QR| = \frac{a\sqrt{2}}{2}$.
Let's find the angle $\\angle PQR$. $\\vec{QP} = P - Q$. Using A=(0,0,0), B=(a,0,0), D=(0,a,0), F=(a,0,a).
P = (0, a/2, 0), Q = (a/2, 0, 0), R = (a, 0, a/2).
$\\vec{QP} = (-a/2, a/2, 0)$. $\\vec{QR} = (a/2, 0, a/2)$.
$\\vec{QP} \cdot \\vec{QR} = (-a/2)(a/2) + (a/2)(0) + (0)(a/2) = -a^2/4$.
$\\cos(\angle PQR) = \frac{\vec{QP} \cdot \\vec{QR}}{|QP||QR|} = \frac{-a^2/4}{(\frac{a\sqrt{2}}{2})(\frac{a\sqrt{2}}{2})} = \frac{-a^2/4}{a^2/2} = -1/2$.
So $\\angle PQR = 120^\circ$.
The area of the rhombus is $|PQ| \times |QR| \times \sin(\angle PQR) = \frac{a\sqrt{2}}{2} \times \frac{a\sqrt{2}}{2} \times \sin(120^\circ) = \frac{a^2 \times 2}{4} \times \frac{\sqrt{3}}{2} = \frac{a^2}{2} \times \frac{\sqrt{3}}{2} = \frac{a^2 \sqrt{3}}{4}$.
Given $a=9$ cm, Area $= \frac{9^2 \sqrt{3}}{4} = \frac{81 \sqrt{3}}{4}$ cm$^2$.
Alternative method: diagonals. We found $PR = \frac{a\sqrt{6}}{2}$. Need QS.
QS connects Q=(a/2,0,0) to S. S is on EH. EH connects E=(0,0,a) to H=(0,a,a). Line EH is $(0, y, a)$ for $0 	ext{ le } y 	ext{ le } a$.
Plane PQR: $x+y-z = a/2$. Substitute S=(0, y, a): $0 + y - a = a/2 => y = 3a/2$. This point is not on EH.
My coordinate system or interpretation of the intersection is wrong.
Let's assume the shape is indeed a rhombus and the calculation of sides is correct.
$PQ = \frac{9\sqrt{2}}{2}$. $QR = \frac{9\sqrt{2}}{2}$.
If the shape is a rhombus, the area can also be calculated using the diagonals. Let's find the length of the diagonals PR and QS.
P=(0, 9/2, 0), Q=(9/2, 0, 0), R=(9, 0, 9/2).
$PR = |R-P| = |(9, -9/2, 9/2)| = \sqrt{9^2 + (-9/2)^2 + (9/2)^2} = \sqrt{81 + 81/4 + 81/4} = \sqrt{81 + 81/2} = \sqrt{243/2} = 9\sqrt{3/2} = \frac{9\sqrt{6}}{2}$.
Let S be the intersection on EH. Line PQ equation: $(a/2, 0, 0) + t(-a/2, a/2, 0)$.
EH line: $(0, y, a)$, $0 	ext{ le } y 	ext{ le } a$. So $S=(0, y_s, a)$.
For S to be on the line PQ extended, $z$ coordinate must match. But PQ has $z=0$. EH has $z=a$. This means PQ does not intersect EH in this coordinate system.
Let's assume the problem implies that the plane PQR cuts the cube to form a closed polygon. P on AD, Q on AB, R on BF. The plane must exit the cube through other edges.
Consider the symmetry. P, Q, R are midpoints. The plane is likely to cut symmetrically.
If the plane cuts EH at S and CG at T, then PQRS and QRTX, etc. will be formed.
Let's assume the intersection is a rhombus PQRS where S is on EH.
We calculated $|PQ| = |QR| = \frac{a\sqrt{2}}{2}$.
If PQRS is a rhombus, we need its diagonals or angle.
Let's verify the angle PQR again. $\\vec{QP} = (-a/2, a/2, 0)$. $\\vec{QR} = (a/2, 0, a/2)$. $\cos(\angle PQR) = -1/2$. So $\\angle PQR = 120^\circ$.
Area = $(\frac{a\sqrt{2}}{2})^2 \sin(120^\circ) = \frac{a^2}{2} \frac{\sqrt{3}}{2} = \frac{a^2 \sqrt{3}}{4}$.
With $a=9$, Area = $\frac{81 \sqrt{3}}{4}$ cm$^2$.
If the shape is a square, the angle would be 90 degrees.
Let's check if the diagonals are equal. $PR = \frac{a\sqrt{6}}{2}$.
Let's find QS. Q = (a/2, 0, 0). S = (0, y_s, a).
S lies on the plane $x+y-z=a/2$. So $0 + y_s - a = a/2 => y_s = 3a/2$. S = (0, 3a/2, a). This point is not on EH.
There might be a misunderstanding of how the plane intersects the cube.
Let's assume the shape is a rhombus and the side length is $s = \frac{a\sqrt{2}}{2}$.
Let's reconsider the shape. P on AD, Q on AB, R on BF. The plane passes through these points. This plane will intersect edges EH and CG.
Let the intersection with EH be S and with CG be T. Then the intersection is PQRT.
$PQ = \frac{a\sqrt{2}}{2}$. $QR = \frac{a\sqrt{2}}{2}$. $RT$ is parallel to $QC$. $QC$ connects Q(mid AB) to C(corner). $QC = \sqrt{BC^2 + QB^2} = \sqrt{a^2 + (a/2)^2} = \sqrt{5a^2/4} = \frac{a\sqrt{5}}{2}$.
$PT$ connects P(mid AD) to T(on CG). T is midpoint of CG? Not specified.
Let's assume the intersection is PQRS as described in many geometry problems of this type, where S is on EH.
If PQRS is a rhombus, side length is $\frac{a\sqrt{2}}{2}$. Area = $\frac{a^2\sqrt{3}}{4}$.
Let's verify the angle calculation. $a=9$. $PQ = \frac{9\sqrt{2}}{2}$. $QR = \frac{9\sqrt{2}}{2}$. $\angle PQR = 120^\circ$. This implies the rhombus is 'wide'.
Let's check the diagonals again. $PR = \frac{9\sqrt{6}}{2}$.
Let S be on EH. S = (0, y, a). Plane equation $x+y-z=a/2$. So $0+y-a = a/2 => y = 3a/2$. S = (0, 3a/2, a). This point is not on EH.
Is the shape a rectangle or a square?
If it's a rectangle, the angle PQR must be 90 degrees. $\cos(\angle PQR) = 0$. But we got -1/2.
Let's use a different approach for the area. Area = base * height. Or using coordinates of vertices.
P=(0, a/2, 0), Q=(a/2, 0, 0), R=(a, 0, a/2). Let S be on EH. EH: $(0,y,a)$ for $0 	ext{ le } y 	ext{ le } a$. S = $(0,y_s,a)$.
Vector PS must be parallel to QR. $\\vec{PS} = S-P = (0, y_s - a/2, a)$. $\\vec{QR} = (a/2, 0, a/2)$. Not parallel.
This means my initial assumption about the shape PQRS and the location of S might be incorrect based on the given points.
Let's assume the points P, Q, R define a plane. The intersection of this plane with the cube is the shape.
P(mid AD), Q(mid AB), R(mid BF).
Face ABCD: intersects at PQ.
Face ABFE: intersects at QR.
Face BCGF: intersects at R and some point T on CG.
Face ADHE: intersects at P and some point S on EH.
Face CDHG: intersects at some point U on DH and V on HG.
Face EFGH: intersects at S on EH and some point W on FG.
Let's use the property that opposite sides of the intersection polygon are parallel. PQ is parallel to FH. So the intersection on EFGH must be a line parallel to FH passing through S. Let this line intersect FG at S'. So SS' is parallel to FH.
Also QR is parallel to AF. So the intersection on CDHG must be a line parallel to AF. This seems wrong.
Let's assume the intersection is a quadrilateral PQRS, with P on AD, Q on AB, R on BF, and S on EH.
$PQ = \frac{a\sqrt{2}}{2}$. $QR = \frac{a\sqrt{2}}{2}$. $RS = ?$ $SP = ?$ 
Since PQRS is planar, $\\vec{PS}$ must be parallel to $\\vec{QR}$, and $\\vec{PQ}$ must be parallel to $\\vec{SR}$.
$\\vec{QR} = (a/2, 0, a/2)$. So $\\vec{PS}$ must be parallel to $(1, 0, 1)$.
$\\vec{PQ} = (-a/2, a/2, 0)$. So $\\vec{SR}$ must be parallel to $(-1, 1, 0)$.
Let S = (0, y, a) on EH. $\\vec{PS} = (0, y-a/2, a)$. This is not parallel to $(1, 0, 1)$.
There is a fundamental issue in my understanding or the coordinate setup.
Let's use the property that the cross section is a polygon whose sides are intersections of the plane with the faces of the cube.
P on AD, Q on AB, R on BF. The plane cuts faces ABCD (segment PQ), ABFE (segment QR), BCGF (segment RT, T on CG), ADHE (segment PS, S on EH), EFGH (segment ST), CDHG (segment PT? No, RS).
So the intersection is PQRS, where P on AD, Q on AB, R on BF, S on EH.
$PQ = \frac{a\sqrt{2}}{2}$. $QR = \frac{a\sqrt{2}}{2}$.
For PQRS to be planar, $\\vec{PQ}$ must be parallel to $\\vec{SR}$ and $\\vec{PS}$ must be parallel to $\\vec{QR}$.
$\\vec{QR} = (a/2, 0, a/2)$ (using A=(0,0,0), B=(a,0,0), D=(0,a,0), E=(0,0,a), F=(a,0,a), H=(0,a,a)).
Let S = (0, y, a) on EH. $\\vec{PS} = S-P = (0, y-a/2, a)$. For PS || QR, we need $\\vec{PS} = k \\vec{QR}$.
$(0, y-a/2, a) = k(a/2, 0, a/2)$.
$0 = k a/2 => k=0$. But then $a = k a/2 = 0$, contradiction.
Let's try different coordinates for EH. E=(0,0,a), H=(0,a,a). EH is $(0, y, a)$ for $y \	ext{ from } 0 	ext{ to } a$.
Let's retry the vector for QR. R=(a, 0, a/2), B=(a,0,0), F=(a,0,a). Q=(a/2, 0, 0). R=(a, 0, a/2). $\\vec{QR} = R-Q = (a/2, 0, a/2)$. Correct.
P=(0, a/2, 0). S=(0, y, a). $\\vec{PS} = S-P = (0, y-a/2, a)$.
For PS || QR, we need $(0, y-a/2, a)$ parallel to $(a/2, 0, a/2)$. This requires the first component to be zero, meaning $a/2=0$ which is impossible.
This means the shape is not a simple PQRS where S is on EH.
Let's reconsider the problem statement and standard geometry of cube cross-sections.
P, Q, R are midpoints. The plane PQR is defined.
Let's use the properties of the plane.
Plane PQR. Normal vector $\vec{n} = \vec{QP} \times \vec{QR} = (-a/2, a/2, 0) \times (a/2, 0, a/2) = (a^2/4, a^2/4, -a^2/4)$. Normal $(1, 1, -1)$.
Equation of plane: $1(x-a/2) + 1(y-0) - 1(z-0) = 0 
=> x + y - z = a/2$.
Now find intersections with edges.
AB: $y=0, z=0$. $x = a/2$. Point is $(a/2, 0, 0)$, which is Q.
AD: $x=0, z=0$. $y = a/2$. Point is $(0, a/2, 0)$, which is P.
BF: $x=a, y=0$. $a + 0 - z = a/2 => z = a/2$. Point is $(a, 0, a/2)$, which is R.
EH: $x=0, z=a$. $0 + y - a = a/2 => y = 3a/2$. Point is $(0, 3a/2, a)$. This point is outside the segment EH (y from 0 to a).
This implies the plane PQR does not intersect EH in the segment EH.
Let's check intersections with other edges.
FG: $x=a, z=a$. $a + y - a = a/2 => y = a/2$. Point is $(a, a/2, a)$. Let's call this S.
CG: $x=a, y=a$. $a + a - z = a/2 => z = 3a/2$. Point outside CG.
DH: $x=0, y=a$. $0 + a - z = a/2 => z = a/2$. Point is $(0, a, a/2)$. Let's call this T.
GH: $y=a, z=a$. $x + a - a = a/2 => x = a/2$. Point is $(a/2, a, a)$.
So the intersection points are Q on AB, P on AD, R on BF, S on FG, T on DH. These are 5 points. The intersection is a pentagon.
P=(0, a/2, 0)
Q=(a/2, 0, 0)
R=(a, 0, a/2)
S=(a, a/2, a)
T=(0, a, a/2)
Let's check if these points are coplanar with the plane $x+y-z=a/2$.
P: $0+a/2-0 = a/2$. Yes.
Q: $a/2+0-0 = a/2$. Yes.
R: $a+0-a/2 = a/2$. Yes.
S: $a+a/2-a = a/2$. Yes.
T: $0+a-a/2 = a/2$. Yes.
So the intersection is a pentagon PQRST.
b. The shape of the cross-section is a pentagon PQRST.
Proof: The plane defined by P, Q, R intersects the edges AB, AD, BF, FG, DH at points Q, P, R, S, T respectively. These five points form the vertices of the pentagon.
c. Calculate the area of the pentagon PQRST.
We can decompose the pentagon into triangles or use the coordinates with the shoelace formula.
Area = $\frac{1}{2} | (x_P y_Q + x_Q y_R + x_R y_S + x_S y_T + x_T y_P) - (y_P x_Q + y_Q x_R + y_R x_S + y_S x_T + y_T x_P) |$
P=(0, a/2, 0), Q=(a/2, 0, 0), R=(a, 0, a/2), S=(a, a/2, a), T=(0, a, a/2).
Area = $\frac{1}{2} | (0 \times 0 + a/2 \times 0 + a \times a/2 + a \times a + 0 \times a/2) - (a/2 \times a/2 + 0 \times a + 0 \times a + a/2 \times 0 + a \times 0) |$
Area = $\frac{1}{2} | (0 + 0 + a^2/2 + a^2 + 0) - (a^2/4 + 0 + 0 + 0 + 0) |$
Area = $\frac{1}{2} | (3a^2/2) - (a^2/4) | = \frac{1}{2} | (6a^2/4) - (a^2/4) | = \frac{1}{2} | 5a^2/4 | = \frac{5a^2}{8}$.
With $a=9$, Area $= \frac{5 \times 9^2}{8} = \frac{5 \times 81}{8} = \frac{405}{8}$ cm$^2$. 
Let's recheck the calculation of the shape. The problem statement says 'Lukislah irisan kubus dengan bidang PQR.' P, Q, R are given points. A plane is defined by these three points. The intersection of this plane with the cube is the cross-section.
Let's visualize the points. P is midpoint of AD. Q is midpoint of AB. R is midpoint of BF. Cube edges a=9.
P=(0, 4.5, 0). Q=(4.5, 0, 0). R=(9, 0, 4.5).
Plane equation: $x+y-z = 4.5$. 
Intersection with EH (x=0, z=9): $0+y-9 = 4.5 => y=13.5$. Outside EH.
Intersection with FG (x=9, z=9): $9+y-9 = 4.5 => y=4.5$. Point S=(9, 4.5, 9).
Intersection with DH (x=0, y=9): $0+9-z = 4.5 => z=4.5$. Point T=(0, 9, 4.5).
So the vertices are Q(4.5, 0, 0), P(0, 4.5, 0), T(0, 9, 4.5), S(9, 4.5, 9), R(9, 0, 4.5).
This is indeed a pentagon.
b. The shape is a pentagon. Proof: The plane intersects 5 edges of the cube.
c. Area of pentagon PQRST with a=9.
P=(0, 4.5, 0), Q=(4.5, 0, 0), R=(9, 0, 4.5), S=(9, 4.5, 9), T=(0, 9, 4.5).
Area = $\frac{5a^2}{8} = \frac{5 \times 9^2}{8} = \frac{405}{8}$.
Let's write the answer. For b, 'irisan kubus' means the shape formed by the intersection of the plane PQR with the cube. P is on AD, Q on AB, R on BF. The plane defined by P, Q, R will intersect other edges of the cube. As calculated, it intersects EH at (0, 13.5, 9) which is outside EH. It intersects FG at (9, 4.5, 9), let's call it S. It intersects DH at (0, 9, 4.5), let's call it T. It intersects GH at (4.5, 9, 9). Let's call it U. It intersects CG at (9, 9, 4.5). Let's call it V.
Let's recheck the calculation of intersections.
Plane: $x+y-z = a/2$. $a=9 => x+y-z=4.5$.
AB: $y=0, z=0 => x=4.5$. Q=(4.5, 0, 0).
AD: $x=0, z=0 => y=4.5$. P=(0, 4.5, 0).
BF: $x=9, y=0 => 9-z=4.5 => z=4.5$. R=(9, 0, 4.5).
FG: $x=9, z=9 => 9+y-9=4.5 => y=4.5$. S=(9, 4.5, 9).
DH: $x=0, y=9 => 9-z=4.5 => z=4.5$. T=(0, 9, 4.5).
GH: $y=9, z=9 => x+9-9=4.5 => x=4.5$. U=(4.5, 9, 9).
So the plane intersects 6 edges: AB, AD, BF, FG, DH, GH.
The intersection polygon has 6 vertices: P, Q, R, S, U, T? No, it should be a simple polygon. The plane cuts through the cube.
The sequence of vertices along the boundary of the intersection must be consecutive edges. Let's trace.
Start at P on AD. Then to Q on AB. Then to R on BF. Then S on FG. Then U on GH. Then T on DH. Then back to P on AD.
So the intersection is a hexagon PQRSTU.
b. The shape of the cross-section is a hexagon PQRSTU.
Proof: The plane intersects six edges of the cube at points P, Q, R, S, U, T.
c. Calculate the area of the hexagon PQRSTU.
P=(0, 4.5, 0), Q=(4.5, 0, 0), R=(9, 0, 4.5), S=(9, 4.5, 9), U=(4.5, 9, 9), T=(0, 9, 4.5).
Shoelace formula for hexagon:
Area = $\frac{1}{2} | (x_P y_Q + x_Q y_R + x_R y_S + x_S y_U + x_U y_T + x_T y_P) - (y_P x_Q + y_Q x_R + y_R x_S + y_S x_U + y_U x_T + y_T x_P) |$
Area = $\frac{1}{2} | (0*0 + 4.5*0 + 9*4.5 + 9*9 + 4.5*4.5 + 0*0) - (4.5*4.5 + 0*9 + 0*9 + 4.5*4.5 + 9*0 + 4.5*0) |$
Area = $\frac{1}{2} | (0 + 0 + 40.5 + 81 + 20.25 + 0) - (20.25 + 0 + 0 + 20.25 + 0 + 0) |$
Area = $\frac{1}{2} | 141.75 - 40.5 | = \frac{1}{2} | 101.25 | = 50.625$.
Let's use $a=9$. Area = $\frac{1}{2} | (0 + 0 + a(a/2) + a(a/2) + (a/2)(a/2) + 0) - ( (a/2)(a/2) + 0 + 0 + (a/2)(a/2) + 0 + 0) |$
Area = $\frac{1}{2} | (a^2/2 + a^2/2 + a^2/4) - (a^2/4 + a^2/4) |$
Area = $\frac{1}{2} | (a^2 + a^2/4) - (a^2/2) | = \frac{1}{2} | 5a^2/4 - 2a^2/4 | = \frac{1}{2} | 3a^2/4 | = \frac{3a^2}{8}$.
With $a=9$, Area $= \frac{3 \times 81}{8} = \frac{243}{8} = 30.375$. My calculation is different.
Let's recheck shoelace formula calculation.
P=(0, a/2), Q=(a/2, 0), R=(a, 0), S=(a, a/2), U=(a/2, a), T=(0, a).
Area = $\frac{1}{2} | (0*0 + a/2*0 + a*a/2 + a*a + a/2*a + 0*a/2) - (a/2*a/2 + 0*a + 0*a + a/2*a/2 + a*0 + a*0) |$
Area = $\frac{1}{2} | (0 + 0 + a^2/2 + a^2 + a^2/2) - (a^2/4 + 0 + 0 + a^2/4 + 0 + 0) |$
Area = $\frac{1}{2} | (2a^2) - (a^2/2) | = \frac{1}{2} | 3a^2/2 | = \frac{3a^2}{4}$.
With $a=9$, Area $= \frac{3 \times 81}{4} = \frac{243}{4} = 60.75$. Still different.
Let's check the plane intersection again. P(mid AD), Q(mid AB), R(mid BF).
Plane: $x+y-z=a/2$.
Vertices: P(0, a/2, 0), Q(a/2, 0, 0), R(a, 0, a/2).
Intersects EH (x=0, z=a) at $y=3a/2$ (outside).
Intersects FG (x=a, z=a) at $y=a/2$. S=(a, a/2, a).
Intersects GH (y=a, z=a) at $x=a/2$. U=(a/2, a, a).
Intersects DH (x=0, y=a) at $z=a/2$. T=(0, a, a/2).
So the vertices are P, Q, R, S, U, T in order.
P=(0, a/2, 0)
Q=(a/2, 0, 0)
R=(a, 0, a/2)
S=(a, a/2, a)
U=(a/2, a, a)
T=(0, a, a/2)
Side lengths:
PQ = $\frac{a\sqrt{2}}{2}$
QR = $\frac{a\sqrt{2}}{2}$
RS = $|S-R| = |(0, a/2, a/2)| = \sqrt{0 + a^2/4 + a^2/4} = \frac{a\sqrt{2}}{2}$
SU = $|U-S| = |(-a/2, a/2, 0)| = \sqrt{a^2/4 + a^2/4} = \frac{a\sqrt{2}}{2}$
UT = $|T-U| = |(-a/2, 0, -a/2)| = \sqrt{a^2/4 + 0 + a^2/4} = \frac{a\sqrt{2}}{2}$
TP = $|P-T| = |(0, -a/2, -a/2)| = \sqrt{0 + a^2/4 + a^2/4} = \frac{a\sqrt{2}}{2}$
All sides are equal. It's a hexagon with equal sides.
Area calculation using shoelace formula with vertices (x,y) pairs in 2D projection. Let's project onto xy plane.
P'=(0, a/2), Q'=(a/2, 0), R'=(a, 0), S'=(a, a/2), U'=(a/2, a), T'=(0, a).
Area = $\frac{1}{2} | (0*0 + a/2*0 + a*a/2 + a*a + a/2*a + 0*a/2) - (a/2*a/2 + 0*a + 0*a + a/2*a/2 + a*0 + a*0) |$
Area = $\frac{1}{2} | (a^2/2 + a^2 + a^2/2) - (a^2/4 + a^2/4) | = \frac{1}{2} | 2a^2 - a^2/2 | = \frac{1}{2} | 3a^2/2 | = \frac{3a^2}{4}$.
With $a=9$, Area = $\frac{3 * 81}{4} = \frac{243}{4} = 60.75$. This should be the projected area.
Area of hexagon = $\frac{3a^2\sqrt{3}}{4}$ if it were regular. But it's not regular.
Let's decompose the hexagon. It can be seen as a rectangle with two triangles removed, or a central rectangle with two trapezoids added.
Consider the projection onto the xy-plane. It's a shape with vertices (0, a/2), (a/2, 0), (a, 0), (a, a/2), (a/2, a), (0, a).
This shape can be seen as a square of side 'a' with two corners cut off. The corners cut off are triangles with vertices (0,0), (a/2,0), (0,a/2) and (a,a), (a/2,a), (a,a/2).
Area of the square is $a^2$. Area of each cut triangle is $\frac{1}{2} (a/2)(a/2) = a^2/8$.
So projected area is $a^2 - 2(a^2/8) = a^2 - a^2/4 = 3a^2/4$. This matches.
Now, this is the projected area. The actual area of the planar hexagon is Area_projected / cos(theta), where theta is the angle between the plane and the xy-plane.
The normal to the plane is $(1, 1, -1)$. The normal to the xy-plane is $(0, 0, 1)$.
$\\cos(\theta) = \frac{|(1, 1, -1) \cdot (0, 0, 1)|}{|(1, 1, -1)| |(0, 0, 1)|} = \frac{|-1|}{\sqrt{1^2+1^2+(-1)^2} \times 1} = \frac{1}{\sqrt{3}}$.
So, Area = Area_projected / cos(theta) = $(3a^2/4) / (1/\sqrt{3}) = \frac{3a^2 \sqrt{3}}{4}$.
With $a=9$, Area $= \frac{3 \times 9^2 \sqrt{3}}{4} = \frac{3 \times 81 \sqrt{3}}{4} = \frac{243 \sqrt{3}}{4}$ cm$^2$.
This seems correct for a hexagon with equal sides. Let's recheck side lengths. They are all $\frac{a\sqrt{2}}{2}$. So it is an equilateral hexagon.
Let's check the angles. Interior angles.
$\\vec{QP} = (-a/2, a/2, 0)$. $\\vec{QR} = (a/2, 0, a/2)$. Angle PQR is 120 deg.
$\\vec{RQ} = (-a/2, 0, -a/2)$. $\\vec{RS} = (0, a/2, a/2)$. $\\vec{RQ} \cdot \\vec{RS} = 0 + 0 - a^2/4 = -a^2/4$.
$|RQ| = \frac{a\sqrt{2}}{2}$. $|RS| = \frac{a\sqrt{2}}{2}$. $\\cos(\angle QRS) = \frac{-a^2/4}{a^2/2} = -1/2$. Angle QRS = 120 deg.
It seems all interior angles are 120 degrees. So it's a regular hexagon in terms of angles. And sides are equal.
So the area formula $\frac{3a^2\sqrt{3}}{4}$ is correct.
With $a=9$, Area = $\frac{243\sqrt{3}}{4}$ cm$^2$.