PQ and PR are two equal chords of a circle with center O.. If PQ = 5 cm and the radius of the circle is 6 cm, find the length of the chord QR
Answers
The length of the chord QR of the circle with centre O and PQ and PR as equal chords is 9.08 cm.
Step-by-step explanation:
It is given that,
A circle with centre O and radius 6 cm.
∆ PQR is inscribed inside the circle such that PQ = PR = 5 cm
Step 1:
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 of a circle, therefore, the centre of the circle O lies on the bisector of ∠QPR.
⇒ OP is the bisector of ∠QPR
Let’s join the points Q & R intersecting OP at M.
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.
∴ The ratio in which M divides QR = 5 : 5 = 1 : 1
i.e., M is mid-point of QR ∴ RM = QM …… (i)
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
Step 2:
Now,
In right-angled triangle ΔPQM, by applying the Pythagoras theorem, we get
PQ² = PM² + QM²
⇒ QM² = PQ² – PM²
⇒ QM² = 5² - PM² ............. (ii)
And,
In right-angled Δ OQM, by applying the Pythagoras theorem, we get
OQ² = OM² + QM²
⇒ OQ² = (OP - PM)² + QM²
⇒ 6² = (6 - PM)² + QM²
⇒ QM² = 36 - (6 - PM)² ........... (iii)
Equating (ii) and (iii), we get
5² - PM² = 36 - (6 - PM)²
⇒ 25 – PM² = 36 – (36 – 12PM + PM²)
⇒ 25 – PM² = 36 – 36 + 12PM - PM²
⇒ 25 = 12 PM
⇒ PM = 25/12
⇒ PM = 2.08 cm
Substituting the value of PM in (ii), we get
QM² = 5² - (2.08)²
⇒ QM = √20.67
⇒ QM = 4.54 cm ….. (iv)
Therefore, from (i) and (iv), 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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