Question

In a circle, PQ is a diameter and AB is a chord. PQ I AB and PQ intersects AB at K. If PQ = 26 cm and KQ = 8 cm. What is the length of AB in cm?​

Answer 1

Required Answer:-

Provided informations:

• PQ is the diameter and AB is the chord.
• PQ is perpendicular to AB.
• PQ intersects AB at K.
• PQ = 26 cm and KQ = 8 cm.

To find: AB's length in cm.

Solution:

Refer to the attachment$.$

Length of the diameter PQ = 26 cm.

Then, radius PO = OQ = 13 cm.

Given, KQ = 8 cm.

And here,

=> OQ = OK + KQ

=> 13 cm = OK + 8 cm

=> OK = 5 cm

We know that,

If a radius is perpendicular to a chord, then it bisects the chord. That means, OQ radius is bisecting chord AB at K. Hence, AK = KB. And angle AKO and angle BKO = 90°

Now In ∆AKO,

• AO = 13 cm (radius of circle)
• OK = 5 cm.

By pythagoras theorem,

=> AO² = AK² + OK²

=> AK = √13² - 5²

=> AK = 12 cm.

Then,

=> AB = 2 × 12 cm (because, AB = 2 AK)

=> AB = 24 cm (ans)

Answer 2

Given :-

In a circle, PQ is the diameter and AB is a chord. PQ║AB and PQ intersects AB at K. If PQ = 26 cm and KQ = 8 cm

To Find :-

Length of AB

Solution :-

It is given that

PQ ║ AB

PQ = Diameter

AB = Chord

PQ = 26 cm

KQ = 8 cm

We know that

$\sf Radius=\dfrac{Diameter}{2}$

$\sf OQ=\dfrac{PQ}{2}$

$\sf OQ=\dfrac{26}{2}$

$\sf OQ=13\;cm$

OQ + PO = PQ

13 + PO = 26

PO = 26 - 13

PO = 13 cm

Here, We concluded radius of circle are equal

So, OA = OB = PO  [Radius of cricle are equal]

OA = OB = 13

Now,

OK + KQ = OQ

OK + 5 = 13

OK = 13 - 5

OK = 5 cm

Now

In ΔAOK

OA² = AK² + OK²

(13)² = AK² + (5)²

169 = AK² + 25

169 - 25 = AK²

144 = AK²

√(144) = AK

12 = AK

In ΔBOK

OB² = BK² + OK²

(13)² = BK² + (5)²

169 = BK² + 25

169 - 25 = BK²

144 = BK²

√(144) = BK

12 = BK

Now

AB = AK + BK

AB = 12 + 12

AB = 24 cm

Hence,

Length of AB is 24 cm.