A perpendicular DK = 10cm is drawn to its plane through the vertex of the square ABCD. Angle between planes ABC
A perpendicular DK = 10cm is drawn to its plane through the vertex of the square ABCD. Angle between planes ABC and KBC = 45 degrees Find the area of square ABCD and 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, in which, according to the rule of three perpendiculars, the angle BCK = 90, CB = 10 cm, and CK = 10 * √2.
Then the area of the triangle BCK will be equal to:
Svsk = BC * CK / 2 = 10 * 10 * √2 / 2 = 50 * √2 cm2.
Answer: Savsd = 100 cm2, Svsk = 50 * √2 cm2.