Question

Two concentric circles are of radii 5 cm ahd 3 cm. Find the length of the chord of the larger
circle which touches the smaller circle.

Answer 1

Answer:

Given : Two concentric circles of radii 5 cm and 3 cm.

To Find : Length of chord AB

Solution : In right angled △ACO,

(AO)² = (AC)² + (CO)²

(AC)² = (AO)² - (CO)²

(AC)² = (5)² - (3)²

(AC)² = 25 - 9

(AC)² = 16

AC = √16

AC = 4 cm

Perpendicular from the center of the circle to the chord divides the chord into two equal parts,

∴ AC = CB

AB = AC + CB

AB = AC + AC

AB = 2AC

AB = 2(4)

AB = 8 cm

the length of the chord of the larger circle which touches the smaller circle is 8 cm.

Answer 2

Question :

• Two concentric circles are of radii 5 cm ahd 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer :

• $\implies \underline{\boxed{\pink{\sf PQ = 8 cm }}}$

Solution :

Understanding the Concept !

We have to find the length of chord (PQ) of larger circle. ∆OMP is right angled triangle. So, we can find the length of PM (base) Using Pythagoras theorem . As OM is perpendicular on PQ . So, length of PM = MQ. Hence, PQ is double of PM.

[REFER TO THE ATTACHMENT]

Let's solve,

∆OMP is right angled triangle.

So,

⇨ PO² = PM² + OM²

⇨ 5² = PM² + 3²

⇨ 25 = PM² + 9

⇨ PM² = 25 - 9

⇨ PM² = 16

$\implies \overline{\boxed{\pink{\sf PM = 4 cm }}}$

Now,

OM is perpendicular on PQ. Means that PM = MQ.

So,

⇨ PQ = PM + MQ.

⇨ PQ = 4 + 4

$\implies \underline{\boxed{\pink{\sf PQ = 8 cm }}}$

Hence ,

The length of the chord of the larger circle which touches the smaller circle is 8 cm.

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