AC and BC are two equal chords of a circle with diameter AB forming a ΔABC as shown in the figure. If the radius of the circle is 5 cm. find the length of the equal chords.
Given radius AO = 5 cm, then diameter AB =10 cm
In ΔACB
∠ACB=90∘ (Angle subtended by diameter AB on circumference)
∠A=∠B (ABC is an isosceles traingle, AC = BC)
∠A+∠B+90∘=180∘
x+x+90∘=180∘⇒x=45∘
Now, angle of triangle ABC are 45∘,45∘,90∘
So, sides AC, BC, AB will be in the ratio 1:1:√2
The corresponding sides can be calculated as.
45∘45∘90∘1:1:√2x:x:x√2ACCBAB↓↓↓5√25√210
(x√2 = 10, ⇒ x=10√2 = 5√2)
So, the chord AC =CB =5√2cm