Chord AB with a length of 40 cm of a circle with center O is perpendicular

| September 18, 2021 | education

Chord AB with a length of 40 cm of a circle with center O is perpendicular to the diameter DC and is at a distance of 21 cm from the center of the circle. 1 Find the radius of the circle 2 In what ratio does the diameter DC divide the point of its intersection with the chord AB.

Let us draw from point O the radii OA and OB to the ends of the chord. The AOB triangle is isosceles, since ОА = ОВ = R.

The height OH of the triangle AOB is also the median of the triangle, then AH = BH = AB / 2 = 40/2 = 20 cm.

In a right-angled triangle AOH, by the Pythagorean theorem, we determine the length of the hypotenuse OA.

OA ^ 2 = AH ^ 2 + OH ^ 2 = 20 ^ 2 + 21 ^ 2 = 400 + 441 = 841.

ОА = R = 29 cm.

Segment OC = ОА = ОВ = R = 29 cm, then СD = D = 2 * R = 2 * 29 = 58 cm.

The length of the segment CH = OC – OH = 29 – 21 = 8 cm.

The length of the segment is DН = СD – СН = 58 – 8 = 50 cm.

Then CH / DH = 8/50 = 4/25.

Answer: The radius of the circle is 29 cm, the diameter is divided in the ratio of 4/25.



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