In a regular quadrangular prism, each side of the base is 4. Through the diagonal of the base at an angle of 45 degrees to the plane of the base, a plane is drawn that intersects the lateral edge. Find the cross-sectional area.
Then the diagonal ВD = СD * √2 = 4 * √2 cm.
The diagonals of the square intersect at right angles and are divided in half, then OC = BD / 2 = 2 * √2 cm.
The BDK triangle is rectangular and isosceles, since the angle C is straight, and the angle O = 45 by condition. Then CK = OC = 2 * √2 cm.
OK = CK * √2 = 2 * √2 * √2 = 4 cm.
Then Svdk = BD * OK / 2 = 4 * √2 * 4/2 = 8 * √2 cm2.
Answer: The cross-sectional area is 8 * √2 cm2.
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