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

| February 26, 2021 | education

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

In a right-angled triangle ABC, the leg BC lies opposite the angle 30, then BC = AB / 2 = 2/2 = 1 cm. Cos45 = AB / BA1. BA1 = AB / Cos45 = 2 / (√2 / 2) = 4 / √2 = 2 * √2 cm.

Right-angled triangle AB1B is isosceles, since one of its angles is 450, then BB1 = AB = 1 cm.Since the side face of AA1B1B is a rectangle, BA1 = AB1 = 2 * √2 cm.

Triangle BCA1 is rectangular with a right angle C, since the AA1C1C facet is perpendicular to CC1B1B, and CA1 and BC belong to these faces.

Then CA1 ^ 2 = BA1 ^ 2 – BC ^ 2 = 8 – 1 = 7.

CA1 = √7 cm.

Let us determine the area of ​​the triangle A1CB. Sa1sv = BC * CA1 / 2 = 1 * √7 / 2 = √7 / 2 cm2.

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



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