At the base of the straight prism lies a triangle ABC with sides AB = 13, BC = 14, AC = 15.

| June 18, 2021 | education

At the base of the straight prism lies a triangle ABC with sides AB = 13, BC = 14, AC = 15. Side rib AA1 = 28. Point M belongs to AA1 and AM: MA1 = 4: 3. Find the BMC sectional area.

Let’s calculate the length of the segment AM.

The path is the length of the segment A1M = 3 * X cm, then the length of the segment AM = 4 * X cm.

A1M + AM = AA1 = 28 cm.

3 * X + 4 * X = 7 * X = 28 cm.

X = 28/7 = 4.

AM = 4 * 4 = 16 cm.

The side faces of the prism are rectangles, then the triangles AFM and ABM are rectangular, then by the Pythagorean theorem:

CM ^ 2 = AC ^ 2 + AM ^ 2 = 225 + 256 = 481

CM = √481 cm.

BM ^ 2 = AB ^ 2 + AM ^ 2 = 169 + 256 = 425.

ВM = 5 * √17 cm.

In the section of the СВM, we will draw the height of the MH.

Let BH = X cm, then CH = (14 – X) cm.

Then:

MH ^ 2 = MВ ^ 2 – BH ^ 2 = 481 – X ^ 2.

MH ^ 2 = CM ^ 2 – CH ^ 2 = 425 – (14 – X) ^ 2.

481 – X ^ 2 = 425 – 196 + 28 * X – X ^ 2.

28 * X = 252.

X = BH = 252/28 = 9 cm.

MH2 = 481 – 81 = 400.

MH = 20 cm.

Then Ssec = BC * MН / 2 = 14 * 20/2 = 140 cm2.

Answer: The cross-sectional area is 140 cm2.



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