The chord cd is 8 cm long and is perpendicular to the diameter ab and bisects the radius ob

The chord cd is 8 cm long and is perpendicular to the diameter ab and bisects the radius ob. find the perimeter of the triangle cad.

Chord CD and diameter AB meet at point H.
1. Since AB is the diameter of the circle, ∠ACB and ∠ADB are equal to 90 °, since they rest on an arc equal to 180 °.
2. Let’s change the radius OC and OD, then the quadrilateral DOCB is a rhombus, and OB and CD are the diagonals of the rhombus, since they intersect at right angles. The diagonals of the rhombus are halved by the intersection point, which means:
OH = BH = OB / 2;
CH = DH = CD / 2 = 8/2 = 4 cm.
3. Consider △ OCB: OC = OB, CH = 4 cm – height. Since OH = BH, then CH is the median. Thus, CH is both the height and the median, then OC = BC. Since OC = OB = BC, △ OCB is correct.
Since △ OCB is correct, ∠BOC = ∠OCB = ∠CBO = 60 °.
4. Find the degree measure ∠ACH:
∠ACH = ∠ACB – ∠OCB / 2;
∠ACH = 90 ° – 60 ° / 2;
∠ACH = 90 ° – 30 °;
∠ACH = 60 °.
Find the degree measure ∠ADH:
∠ADH = ∠ADB – ∠ODB / 2;
∠ADH = 90 ° – 60 ° / 2;
∠ADH = 90 ° – 30 °;
∠ADH = 60 °.
5. Consider △ DAC: ∠ACH = ∠ADH = 60 °. By the theorem on the sum of the angles of a triangle:
∠ACH + ∠ADH + ∠DAC = 180 °;
60 ° + 60 ° + ∠DAC = 180 °;
∠DAC = 180 ° – 120 °;
∠DAC = 60 °.
∠DAC = ∠ACH = ∠ADH = 60 °, therefore △ DAC is correct: AD = AC = CD = 8 cm.
6. Perimeter △ CAD is equal to:
P = 3a,
where a is the side length of a regular triangle.
P = 3 * 8 = 24 (cm).
Answer: P = 24 cm.