Question

out of two concentric circles, the radius of the outer circle is 25 cm and the chord ac
of length 48 cm is a tangent to the inner circle. Find the radius of the inner circle​

Answer 1

According to question,

Chord AB = 48 cm

OA = 25 cm (given)

OC is perpendicular to AB

Now, AC = CB = 48/2cm =24 cm

In right-angled ∆ OCA,

OC² + CA² = OA² (Pythagoras Theorem)

=> OC² = OA²-CA²

=> OC² = (25)²-(24)²

=> OC² = 625-576

=> OC² = 49

=> OC = 7cm

Ans) Radius of Inner Circle is 7cm

Answer 2

The radius of inner circle=7 cm

Step-by-step explanation:

Radius of outer circle=OA=25 cm

Length of chord=AB=48 cm

We know that

Radius is perpendicular to tangent .

The perpendicular drawn from the center bisect the chord

Therefore,AD=DB=$\frac{48 cm}{2}=24 cm$

In triangle OBA,

By pythagprous theorem

$(Hypotenuse)^2=(base)^2+(perpendicular\;side)^2$

$(OA)^2=(AD)^2+(OD)^2$

Substitute the values

$(25)^2=(24)^2+(OD)^2$

$625=576+(OD)^2$

$OD^2=625-576=49$

$OD=\sqrt{49}=7 cm$

Hence, radius of inner circle=7 cm

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