At the base of the regular triangular prism ABCA1B1C1 lies the triangle ABC, AB = 6.

| May 25, 2021 | education

At the base of the regular triangular prism ABCA1B1C1 lies the triangle ABC, AB = 6. Prism height AA1 = 3. Find the distance between straight line A1B1 and plane ABC1.

On the side face АА1С1С we define the diagonal АС1.

AC1 ^ 2 = AA1 ^ 2 + AC ^ 2 = 9 + 36 = 45.

AC1 = √45 cm.

The side faces of the prism are equal, then BC1 = AC1 = √45 cm, then triangle ABC1 is isosceles. Let’s draw the height C1H of the isosceles triangle ABC1, which is also the median of the triangle. From the right-angled triangle AC1H, we determine the length of the leg C1H.

C1H ^ 2 = A1C ^ 2 – AH ^ 2 = 45 – 9 = 36.

С1Н = 6 cm.

Let’s draw the height С1К of the equilateral triangle А1В1С1. C1K = A1B1 * √3 / 2 = 6 * √3 / 2 = 3 * √3.

The area of ​​the triangle С1КН will be equal to: Sс1кн = С1К * КН / 2 = 3 * √3 * 3/2 = 9 * √3 / 2 cm2.

Also, the area of ​​the triangle С1КН is equal to: Sс1кн = С1Н * КМ / 2 = 6 * КМ / 2 = 9 * √3 / 2.

6 * KM = 9 * √3.

KM = 9 * √3 / 6 = 3 * √3 / 2 = 1.5 * √3 cm.

Answer: From straight line A1B1 to plane ABC1 1.5 * √3 cm.



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