Two parallel chords of length 8cm and 6 cm of a circle are separated by a distance of 1cm. Find the radius of the circle. Plz answer the Question quickly
Answers
Answer:
4cm is the radius of the cirule
If two parallel chords of length 8 cm and 6 cm of a circle are separated by a distance of 1 cm then the radius of the circle is 5 cm.
Step-by-step explanation:
It is given that,
Two chords of length 6 cm & 8 cm of a circle are parallel to each other.
Chords are separated by a distance of 1 cm.
Referring to the figure attached below,
Let’s draw a line OE from the centre O intersecting the chord CD at P, AB at Q and the circle at E i.e., OE is the radius of the circle.
We know that if a radius or a diameter of the circle that is perpendicular to a chord divides the chord into two equal parts and vice versa.
∴ AQ = BQ = ½ * AB = ½ * 6 = 3 cm
And, DP = CP = ½ * 8 = 4 cm
Let the radius of the circle be denoted as OB = OC = “r” cm.
Step 1:
Consider right-angled ∆OPC, by applying Pythagoras theorem,
OC² = OP² + CP²
⇒ r² = OP² + 4²
⇒ OP² = r² – 16
⇒ OP = √[r² – 16]…… (i)
Step 2:
Consider right-angled ∆OQB, by applying Pythagoras theorem,
OB² = OQ² + BQ²
⇒ r² = [OP + 1]² + 3²
⇒ r² = OP² + 2OP + 1 + 9
⇒ OP² + 2 OP = r² – 10
Substituting the value of OP from (i)
⇒ r² – 16 + [2*{√[r² – 16]}] = r² - 10
⇒ 2*{√[r² – 16]} = 6
⇒ √[r² – 16] = 3
Squaring both sides
⇒ r² – 16 = 9
⇒ r² = 25
⇒ r = 5 cm
Thus, the radius of the circle is 5 cm.
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