Question

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Answer 1

Answer:

answer is 7

Step-by-step explanation:

here in triangle PBO(right angled triangle)

PO is perpendicular

PB is base

BO is hypoteus, so by applying Pythagoras theorem:

BO^2=PB^2 +PO^2

PO^2=25-9=16

PO=√16=4cm

Now in triangle QDO (right angled triangle)

DO^2=QO^2 +QD^2

QO^2=25-16=9

QO=√9=3

so the length of PQ=PO+QO=4+3=7

Answer 2

Question :-

In the given figure, O is the centre of circle with radius 5 cm OP ⊥ AB, OQ ⊥ CD, AB || CD, AB = 6 cm, AB = 6 cm and CD = 8 cm. Determine PQ.

Before solving let us do some construction.

Construction :-

Join AO and OD

Solution :-

Consider ΔAOP and ΔDOQ

OP ⊥ AB, OQ ⊥ CD [ Given ]

So, ∠APO = 90°, ∠DQO = 90°

Therefore, ΔAOP and ΔDOQ are Right angled triangles.

In ΔAOP, ΔDOQ

AO = OD = 5 cm [ Radius of the circle ]

AB is a chord drawn to a circle and AB ⊥ OP

Now we need to recall a theorem :

Theorem :- The perpendicular from the centre of a circle to a chord bisects the chords

Therefore, AP = AB / 2 = 6 / 2 = 3 cm

Similarly, QD = CD / 2 = 8 / 2 = 4 cm

In ΔAOP, By pythagoras theorem

AO² = OP² + AP²

⇒ 5² = OP² + 3²

⇒ 25 = OP² + 9

⇒ 25 - 9 = OP²

⇒ 16 = OP²

⇒ OP = √16 = 4 cm

In ΔDOQ, By pythagoras theorem

DO² = OQ² + DQ²

⇒ 5² = OQ² + 4²

⇒ 25 = OQ² + 16

⇒ 25 - 16 = OQ²

⇒ 9 = OQ²

⇒ OQ = √9 = 3 cm

From figure,

PQ = OP + OQ = 4 + 3 = 7 cm.

Hence, the length of PQ is 7 cm.