AB is a chord of a circle with centre O. OP = 3 cm if radius is 5 cm, and OP ⊥ AB, then find
the length of the chord AB
Answers
Given :
AB is a chord of a circle with centre O. OP = 3 cm if radius is 5 cm, and OP ⊥ AB.
To Find :
The length of the chord AB.
Solution :
Analysis :
Here we have to use Pythagoras theorem to find AP & PB and then add have to add up both the bases to find the length of the chord AB.
Explanation :
As AB is the chord. O is the centre of the circle. Since OP ⊥ AB drawn from O as its distance, then ∆OAP and ∆OPB are both right angled triangles.
- Now radius of circle OA = OB = 5 cm. (∵ a circle have same radii)
- OP is common for both the triangles.
In ∆OAP,
By using Pythagoras theorem,
(Hypo)² = (base)² + (side)²
(OA)² = (AP)² + (OP)²
where,
- OA = Hypotenuse = 5 cm
- OP = side = 3 cm
- AP = Base = AP cm
Using the required formula and substituting the required values,
⇒ (5)² = (AP)² + (3)²
⇒ 25 = (AP)² + 9
⇒ 25 - 9 = (AP)²
⇒ 16 = (AP)²
Square rooting both the sides,
⇒ √16 = AP
⇒ √[4 × 4] = AP
⇒ 4 = AP
∴ AP = 4 cm.
Similarly,
In ∆OPB,
By using Pythagoras theorem we will get PB.
∴ PB = 4 cm.
Now,
Chord AB = AP + PB
where,
- AP = 4 cm
- PB = 4 cm
Using the required formula and substituting the required values,
⇒ AB = 4 + 4
⇒ AB = 8
∴ AB = 8 cm.
The length of the chord AB is 8 cm.
(For more reference refer to the attachment)
