#### Question

In the given figure, O is the centre of the circle with radius 5 cm. OP and OQ are perpendicular to AB and CD respectively. AB = 8 cm and CD = 6 cm. determine the length of PQ.

#### Solution

Radius of the circle whose centre is 0 = 5 cm

OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.

Join OA and OC, then OA = OC + 5 cm

Since OP ⊥ AB, P is the midpoint of AB

Similarly, Q is the midpoint of CD

In right ∆OAP,

`OA^2 = OP^2 + AP^2` [Pythagoras Theorem]

⇒ `(5)^2 = OP^2 + (4)^2 [ ∵ AP = PB = 1/2xx 8 = 4 cm]`

⇒ ` 25 = OP^2+16`

⇒ `OP^2 = 25 -16`

⇒ `OP^2 = 9`

⇒OP = 3 cm

Similarly, in right ∆OCQ,

`OC^2 = OQ^2 + CQ^2` [Pythagoras theorem]

⇒`(5)^2 = OQ^2 + (3)^2`

⇒`25 = OQ^2 + 9`

⇒` OQ^2 = 25 - 9`

⇒`OQ^2 = 16`

⇒ OQ = 4cm

Hence, PQ = OP + OQ = 3 + 4 = 7 cm

Is there an error in this question or solution?

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In the Given Figure, O is the Centre of the Circle with Radius 5 Cm. Op and Oq Are Perpendicular to Ab and Cd Respectively. Ab = 8 Cm and Cd = 6 Cm. Determine the Length of Pq. Concept: Chord Properties - the Perpendicular to a Chord from the Center Bisects the Chord (Without Proof).

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