Chord AB, equal to 8 cm, cuts off an arc of 90 degrees from the circle with center O. The diameters AC

| June 20, 2021 | education

Chord AB, equal to 8 cm, cuts off an arc of 90 degrees from the circle with center O. The diameters AC and BD are drawn through the ends of the chord. a) Determine the type of quadrilateral ABCD b) Find the lengths of the diagonals and unknown sides of the quadrilateral c) what angles form the sides of the quadrilateral with a tangent to the circle at point B

Since the chord AB cuts off the arc 90, the central angle AOB is also equal to 90. The diagonals AC and BD, in the center O of the circle are divided in half, and are equal to each other AC = BD, OA = OB = OC = OD.

Since the diagonals intersect at right angles, are equal to each other and are divided in half at the point of intersection, ABCD is a square, and therefore AB = BC = CD = AD = 8 cm.

The diagonal of the square is equal to a * √2, where a is the length of the side of the square.

AC = BD = 8 * √2 cm.

The tangent CM is perpendicular to the diameter BD, and since BD is perpendicular to the AC, the AC is parallel to the CM.

The angle BAC = 45, then the angle KBA = 45, as the angles lying crosswise at the intersection of the parallel KM and AC secant AB. СВM angle = 180 – 90 – 45 = 45.

Answer: ABCD is a square, the length of the diagonals is 8 * √2 cm, the length of the unknown sides is 8 cm, the angle between the tangent and the sides is 45.



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