Consider a Tetrahedron OABC with the following conditions: OA = 4, OB

Berikut ini adalah pertanyaan dari touchmee03 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Consider a Tetrahedron OABC with the following conditions:OA = 4,
OB = AB = BC = 3,
OC = AC = 2√3

Given that P is a point on line BC, and G is geometric center of △OAP,
Find the smallest possible value of PG as P moves along line BC.

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Penjelasan dengan langkah-langkah:

To find the smallest possible value of PG, we can consider the case where point P is at the midpoint of line segment BC. Let the midpoint of line segment BC be M.

We can find the coordinates of point G as the average of the coordinates of points O and P.

G = (1/2) * (O + P)

Plugging in the coordinates of O and P, we get:

G = (1/2) * [(0,0) + (3*sqrt(3)/4,3/4)]

G = (3*sqrt(3)/8,3/8)

Then, we can find the distance between points G and P using the distance formula:

PG = sqrt((Gx - Px)^2 + (Gy - Py)^2)

Plugging in the coordinates of G and P, we get:

PG = sqrt((3sqrt(3)/8 - 3sqrt(3)/4)^2 + (3/8 - 3/4)^2)

PG = sqrt((-3*sqrt(3)/8)^2 + (-3/8)^2)

PG = sqrt((9/8) + (9/64))

PG = sqrt(81/64 + 81/64)

PG = sqrt(162/64)

PG = sqrt(81/32)

PG = 3/4 * sqrt(2)

Therefore, the smallest possible value of PG as P moves along line BC is 3/4 * sqrt(2).

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Last Update: Sun, 09 Apr 23