Question

PQ and PR are two equal chords of a circle with centre O. If
PQ = 5 cm and the radius of the circle is 6 cm, find the length of
the chord QR.

plz solve ​

Answer 1

Given:

PQ and PR are two equal chords of a circle with centre O.

PQ = 5 cm

The radius of the circle is 6 cm

To find:

The length of the chord QR

Solution:

 Construction:- Join O & P such that it intersects QR at M and also join O & Q.

∆ PQR is inscribed inside the circle such that PQ = PR = 5 cm

We know that → the angle bisector of an angle between two equal chords of a circle passes through the centre of the circle.

Here, PQ and PR are given as two equal chords, therefore, the centre of the circle O lies on the bisector of ∠QPR.

⇒ OP is the bisector of ∠QPR

 

Again we know that → the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle

i.e., bisector OP will divide QR in the ratio of PQ : QR = 5 : 5 = 1 : 1  

⇒ M is the mid-point of QR ∴ RM = QM ↔ Equation1

Since we know that → if a line from the centre to chord, divides the chord into two equal parts, then the line joining the chord will be perpendicular to it.

i.e., OM ⊥ QR  

Now,  

In right-angled triangle Δ PQM, using the Pythagoras theorem, we get

PQ² = PM² + QM²

⇒ QM² = PQ² – PM²

⇒ QM² = 5² - PM² ↔ Equation2

And,

In right-angled Δ OQM, using the Pythagoras theorem, we get

OQ² = OM² + QM²

⇒ OQ² = (OP - PM)² + QM²

⇒ 6² = (6 - PM)² + QM²

⇒ QM² = 36 - (6 - PM)² ↔ Equation3

Comparing equation2 and equation3, we get

5² - PM²  = 36 - (6 - PM)²  

⇒ 25 – PM² = 36 – (36 – 12PM + PM²)

⇒ 25 – PM² = 36 – 36 + 12PM - PM²

⇒ 25 = 12 PM

⇒ PM = 2.08 cm

On substituting the value of PM in equation2, we get

QM² = 5² - (2.08)²

⇒ QM = √20.67

⇒ QM = 4.54 cm Equation4

Therefore, from equation1 and equation4, we get

QR = 2 × QM = 2 × 4.54 = 9.08 cm

Thus, the length of the chord QR is → 9.08 cm.

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