At the base of the straight prism ABCA1B1C1 is a triangle ABC, whose angle is C = 90, AB = 2

| April 10, 2021 | education

At the base of the straight prism ABCA1B1C1 is a triangle ABC, whose angle is C = 90, AB = 2, angle BAC = 30, angle B1AB = 45. Find the area of triangle A1CB.

At the base of the prism lies a right-angled triangle ABC in which the leg of the AC is located opposite an angle of 300, then BC = AB / 2 = 2/2 = 1 cm.

The side face of AA1B1B is a rectangle in which the diagonal BA1 divides it into two isosceles, right-angled triangles.

Then AB1 = BA1 = √2 * AB ^ 2 = 2 * √2 cm.

Since the side faces of CC1B1B and AA1C1C are perpendicular, the triangle A1CB is rectangular with a right angle C.

Then A1C ^ 2 = A1B ^ 2 – BC ^ 2 = 8 – 1 = 7.

A1C = √7 cm.

The area of the triangle А1СВ is equal to: Sa1св = ВС * А1С / 2 = 1 * √7 / 2 = √7 / 2 cm2.

Answer: The area of triangle A1CB is equal to √7 / 2 cm2.



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