Diketahui limas beraturan T.ABCD dengan panjang rusuk 6 cm.

2*sqrt(6) cm

Pembahasan

Diketahui limas beraturan T.ABCD dengan panjang rusuk alas (AB = BC = CD = DA) = 6 cm, dan panjang rusuk tegak (TA = TB = TC = TD) juga diasumsikan 6 cm (meskipun tidak disebutkan secara eksplisit, "limas beraturan" biasanya berarti alasnya beraturan dan rusuk tegaknya sama panjang).
P adalah titik tengah CD. Maka, CP = PD = 6/2 = 3 cm.
Kita perlu mencari jarak dari P ke bidang TAB.
Bidang TAB dibentuk oleh titik T, A, dan B.
Kita bisa menggunakan teorema Pythagoras untuk mencari tinggi limas (jarak dari T ke pusat alas ABCD). Misalkan O adalah pusat alas ABCD. Diagonal AC = BD = sqrt(6^2 + 6^2) = sqrt(72) = 6*sqrt(2) cm. Jarak AO = OC = BO = OD = (1/2) * 6*sqrt(2) = 3*sqrt(2) cm.
Tinggi limas TO = sqrt(TA^2 - AO^2) = sqrt(6^2 - (3*sqrt(2))^2) = sqrt(36 - 18) = sqrt(18) = 3*sqrt(2) cm.
Sekarang, kita perlu mencari proyeksi P pada bidang TAB. Karena limas beraturan, kita bisa mempertimbangkan simetri.
Misalkan M adalah titik tengah AB. Maka TM adalah tinggi segitiga TAB, dan TM = sqrt(TA^2 - AM^2) = sqrt(6^2 - 3^2) = sqrt(36 - 9) = sqrt(27) = 3*sqrt(3) cm.
Jarak P ke bidang TAB sama dengan jarak P ke garis TM jika kita memproyeksikan P ke bidang yang melalui M dan sejajar dengan bidang alas dan tegak lurus TM. Namun, ini terlalu rumit.
Cara yang lebih mudah adalah menggunakan konsep vektor atau mencari segitiga siku-siku yang tepat.
Consider the triangle TCD. P is the midpoint of CD. The distance from P to the line segment AB (which lies on the base) is the distance from P to M, where M is the midpoint of AB. PM = BC = 6 cm.
Consider the triangle T M P. We have TM = 3*sqrt(3) cm and PM = 6 cm. The angle TMP is the angle between the face TCD and the base ABCD. No, this is not correct. PM is perpendicular to AB.
Let's rethink. We need the distance from point P to the plane TAB.
Let's set up a coordinate system. Let D = (0,0,0), C = (6,0,0), A = (0,6,0), B = (6,6,0). Then the center of the base O = (3,3,0).
T = (3,3, 3*sqrt(2)).
P = (3,0,0).
The plane TAB passes through A=(0,6,0), B=(6,6,0), T=(3,3, 3*sqrt(2)).
A vector normal to the plane TAB can be found using the cross product of two vectors in the plane, e.g., AB and AT.
AB = B - A = (6, 0, 0)
AT = T - A = (3, -3, 3*sqrt(2))
Normal vector N = AB x AT = (0 * 3*sqrt(2) - 0 * (-3), 0 * 3 - 6 * 3*sqrt(2), 6 * (-3) - 0 * 3) = (0, -18*sqrt(2), -18).
We can simplify the normal vector to N' = (0, sqrt(2), 1).
The equation of the plane TAB is of the form 0x + sqrt(2)y + 1z = d. Using point A(0,6,0): sqrt(2)*6 + 1*0 = d => d = 6*sqrt(2).
So the plane equation is sqrt(2)y + z = 6*sqrt(2).
The distance from point P(3,0,0) to the plane sqrt(2)y + z - 6*sqrt(2) = 0 is given by the formula:
Distance = |Ax0 + By0 + Cz0 + D| / sqrt(A^2 + B^2 + C^2)
Distance = |0*3 + sqrt(2)*0 + 1*0 - 6*sqrt(2)| / sqrt(0^2 + (sqrt(2))^2 + 1^2)
Distance = |-6*sqrt(2)| / sqrt(2 + 1)
Distance = 6*sqrt(2) / sqrt(3)
Distance = 6*sqrt(2)*sqrt(3) / 3
Distance = 6*sqrt(6) / 3
Distance = 2*sqrt(6) cm.
Wait, the question says "limas beraturan T.ABCD dengan panjang rusuk 6 cm". This usually means all edges are 6 cm. Let's assume that.
Then TA=TB=TC=TD=AB=BC=CD=DA=6 cm.
P is the midpoint of CD, so CP = 3 cm.
Let's consider the cross-section through P and perpendicular to CD and AB. Let M be the midpoint of AB. PM = 6 cm.
Consider triangle TCD. TP is the median to CD. Since TCD is an isosceles triangle (TC=TD=6), TP is also the altitude, so TP is perpendicular to CD. TP = sqrt(TC^2 - CP^2) = sqrt(6^2 - 3^2) = sqrt(36 - 9) = sqrt(27) = 3*sqrt(3) cm.
Now consider triangle TPM. PM = 6 cm, TP = 3*sqrt(3) cm. We want the distance from P to the plane TAB. Let H be the projection of P onto the plane TAB. We are looking for the length PH.
Let's use a different approach. Consider the angle between the face TCD and the base ABCD. Let theta be this angle. In triangle T M P, angle TMP is the angle between face TCD and base ABCD. Wait, M is midpoint of AB, P is midpoint of CD. PM is perpendicular to AB and CD. Consider triangle TPM. PM = 6. TM = 3*sqrt(3). TP = 3*sqrt(3).
Triangle TPM is isosceles with TP=TM. Wait, this is not possible if PM=6 and TM=TP=3*sqrt(3) approx 5.19.
Let's re-read the question. "limas beraturan T.ABCD dengan panjang rusuk 6 cm". This typically means the base edges are 6 cm, and the slant edges are also 6 cm. So AB=BC=CD=DA=6 and TA=TB=TC=TD=6.
P is the midpoint of CD, so DP=PC=3.
We want the distance from P to the plane TAB.
Let's find the height of the pyramid. Let O be the center of the base. O is the intersection of diagonals AC and BD. AC = BD = sqrt(6^2+6^2) = 6*sqrt(2). AO = 3*sqrt(2).
Height TO = sqrt(TA^2 - AO^2) = sqrt(6^2 - (3*sqrt(2))^2) = sqrt(36 - 18) = sqrt(18) = 3*sqrt(2).
Let's consider a coordinate system. Let D=(0,0,0), C=(6,0,0), A=(0,6,0), B=(6,6,0). Then T = (3,3, 3*sqrt(2)). P = (3,0,0).
The plane TAB contains points A(0,6,0), B(6,6,0), T(3,3, 3*sqrt(2)).
Vector AB = (6,0,0).
Vector AT = (3, -3, 3*sqrt(2)).
Normal vector N = AB x AT = (0, -18*sqrt(2), -18). We can use N' = (0, sqrt(2), 1).
The equation of the plane TAB is sqrt(2)y + z = d. Using A(0,6,0): sqrt(2)*6 + 0 = d => d = 6*sqrt(2).
Plane equation: sqrt(2)y + z - 6*sqrt(2) = 0.
Point P = (3,0,0).
Distance = |sqrt(2)*0 + 0 - 6*sqrt(2)| / sqrt(0^2 + (sqrt(2))^2 + 1^2) = |-6*sqrt(2)| / sqrt(2+1) = 6*sqrt(2) / sqrt(3) = 2*sqrt(6).
Let's recheck the problem statement. Maybe the pyramid is not regular in terms of slant edges. "limas beraturan T.ABCD" means the base ABCD is a regular polygon (square) and the apex T is directly above the center of the base. It does not necessarily mean the slant edges are equal to the base edges.
However, if it's a regular pyramid with base edge 6, and the slant height is needed, we usually assume slant edges are equal.
Let's assume the problem means all edges are 6 cm.
Let's try geometric approach again. Let M be the midpoint of AB. PM is perpendicular to AB and has length 6. Let N be the midpoint of CD. Wait, P is the midpoint of CD. Let M be the midpoint of AB. Then PM is perpendicular to CD and AB. PM = 6.
Consider triangle TCD. TP is the median to CD. Since TC=TD=6, triangle TCD is isosceles. TP is perpendicular to CD. TP = sqrt(TC^2 - CP^2) = sqrt(6^2 - 3^2) = sqrt(27) = 3*sqrt(3).
Consider triangle TAB. TM is the median to AB. Since TA=TB=6, triangle TAB is isosceles. TM is perpendicular to AB. TM = sqrt(TA^2 - AM^2) = sqrt(6^2 - 3^2) = sqrt(27) = 3*sqrt(3).
Now consider the triangle TPM. PM = 6. TP = 3*sqrt(3). TM = 3*sqrt(3).
We want the distance from P to the plane TAB. Let H be the foot of the perpendicular from P to the plane TAB.
The plane TAB is defined by points T, A, B. TM is in this plane.
Consider the plane containing TPM. The line TM is in the plane TAB. The distance from P to the plane TAB is the height of triangle TPM from P to the base TM (or its extension).
Let's drop a perpendicular from P to the line TM. Let this foot be K. The distance PK is the distance from P to the plane TAB.
In triangle TPM, we have sides TP = 3*sqrt(3), TM = 3*sqrt(3), PM = 6.
Let's find the angle PTM using the Law of Cosines in triangle TPM:
PM^2 = TP^2 + TM^2 - 2 * TP * TM * cos(angle PTM)
6^2 = (3*sqrt(3))^2 + (3*sqrt(3))^2 - 2 * (3*sqrt(3)) * (3*sqrt(3)) * cos(angle PTM)
36 = 27 + 27 - 2 * 27 * cos(angle PTM)
36 = 54 - 54 * cos(angle PTM)
54 * cos(angle PTM) = 54 - 36 = 18
cos(angle PTM) = 18 / 54 = 1/3.
Now, consider the triangle formed by P, T, and the projection of P onto the plane TAB. Let's call the projection K. PK is perpendicular to the plane TAB, so PK is perpendicular to TM.
Consider triangle TPK, where angle PTK is the angle between the face TCD and the plane TAB. This is not straightforward.
Let's use the height from P to TM in triangle TPM. Let K be on TM such that PK is perpendicular to TM. We need to find PK.
Area of triangle TPM = (1/2) * base * height = (1/2) * TM * PK = (1/2) * 3*sqrt(3) * PK.
We can also find the area using Heron's formula. Semiperimeter s = (3*sqrt(3) + 3*sqrt(3) + 6) / 2 = (6*sqrt(3) + 6) / 2 = 3*sqrt(3) + 3.
Area^2 = s(s-a)(s-b)(s-c) = (3*sqrt(3)+3) * (3*sqrt(3)+3 - 3*sqrt(3)) * (3*sqrt(3)+3 - 3*sqrt(3)) * (3*sqrt(3)+3 - 6)
Area^2 = (3*sqrt(3)+3) * (3) * (3) * (3*sqrt(3)-3)
Area^2 = 9 * (3*sqrt(3)+3) * (3*sqrt(3)-3)
Area^2 = 9 * ((3*sqrt(3))^2 - 3^2)
Area^2 = 9 * (27 - 9)
Area^2 = 9 * 18 = 162
Area = sqrt(162) = 9*sqrt(2).
So, (1/2) * 3*sqrt(3) * PK = 9*sqrt(2)
PK = (2 * 9*sqrt(2)) / (3*sqrt(3)) = 18*sqrt(2) / (3*sqrt(3)) = 6*sqrt(2) / sqrt(3)
PK = 6*sqrt(2)*sqrt(3) / 3 = 2*sqrt(6).
This confirms the coordinate geometry result.