Through the vertex of the square ABCD, a perpendicular DK = 10cm is drawn to its plane.

| July 25, 2021 | education

Through the vertex of the square ABCD, a perpendicular DK = 10cm is drawn to its plane. Angle between planes ABC and KBC = 45 degrees. Find the area of the square ABCD and the triangle BCK.

Consider a triangle CKD, in which, by condition, the angle CDK = 90, and the angle DSC = 45, then the angle CKD = 80 – 90-45 = 45.

The angles KCD and CKD are equal, therefore, the rectangle is isosceles and KD = CD = 10 cm.

Since there is a square at the base, AB = BC = CD = AD = 10 cm, and the area of ​​the square is: Savsd = AB * CB = 10 * 10 = 100 cm2.

Let us find the hypotenuse CK of the triangle CKD by the Pythagorean theorem.

CK ^ 2 = CD ^ 2 + CD ^ 2 = 100 + 100 = 200.

CK = √200 = 10 * √2.

Consider a triangle CKB, for which, according to the rule of three perpendiculars, the angle BSC = 90, CB = 10 cm, and CK = 10 * √2.

Then the area of ​​the triangle BSC will be equal to:

Svsk = BC * CK / 2 = 10 * 10 * √2 / 2 = 50 * √2 cm2.

Answer: Savsd = 100 cm2, Svsk = 50 * √2 cm2.



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