Question

two circles touch externally the sum of their areas is 130 pi sq.cm and the distance between their center is 14 cm find the radii of the circle ​

Answer 1

Step-by-step explanation:

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ANSWER

If two circles touch externally, then the distance between their centers is equal to the sum of their radii. Let the radii of the two circles be r

1

cm and r

2

cm respectively.

Let C

1

and C

2

be the centres of the given circles. Then,

C

1

C

2

=r

1

+r

2

[∵C

1

C

2

=14cm (given)]

⇒14=r

1

+r

2

⇒r

1

+r

2

=14 ....(i)

It is given that the sum of the areas of two circles is equal to 130πcm

2

.

∴π(r

1

)

2

+π(r

2

)

2

=130π

⇒(r

2

)

2

+(r

2

)2=130 ...(ii)

Now, (r

1

+r

2

)

2

=(r

1

)

2

+(r

2

)

2

+2r

1

r

2

⇒142=130+2r

1

r

2

[Using (i) and (ii)]

⇒196−130=2r

1

r

2

⇒r

1

r

2

=33 ....(iii)

Now,

(r

1

−r

2

)

2

=(r

1

)

2

+(r

2

)

2

−2r

1

r

2

⇒(r

1

−r

2

)

2

=130−2×33 [Using (ii) and (iii)]

⇒(r

1

−r

2

)

2

=64

⇒r

1

−r

2

=8 ....(iv)

Solving (i) and (iv), we get r

1

=11cm and r

2

=3cm. Hence, the radii of the two circles are 11cm and 3cm.

Answer 2

Answer:

r1 = 11cm, r2 = 3cm

Step-by-step explanation:

Let's assume that the radii of the first circle be r1 and the second circle be r2.

And the centre of first circle be C1 and the second circle be C2.

In the question, it is said that the two circle are touching each other extremely. So it can be said that the distance between their centres will be equal the sum of their radii. It can be written as:

C1 C2 = r1 + r2

14cm = r1 + r2

Now, it is given that the sum if their areas is 130 pi cm².

$\implies \sf \pi r1^2 + \pi r2^2 = 130 \pi$

Pi cancels out from both the sides.

$\implies \sf r1^2 + r2^2 = 130$

Which further simplifies to:

$\sf \implies (r1 +r2)^2$

Now? Yes! To simplify this, we need to apply algebraic identity.

${\bold {\boxed{\bold{(a+b)^2 = a^2 + b^2 + 2ab}}}}$

We know that, r1 +r2 = 14 also that r1² + r2² = 130.

Now,

$\sf \implies (r1+r2) = r1^2 + r2^2 + 2r1r2$

$\implies \sf 14^2 = 130 + 2r1r2$

$\implies \sf 196 - 130 = 2r1r2$

$\sf \implies 66 = 2r1r2$

$\implies \sf r1r2 = 33$

${\bold {\boxed{\bold{(a-b^2 = a^2 + b^2 - 2ab}}}}$

$\implies \sf 130 - 2 \times 33\\\\\implies 64 = (r1-r2)^2 \\\\\implies \sqrt{64} = r1-r2 \\\\\implies 8 = r1-r2$

r1 + r2 = 14, r1 - r2 = 8

r1 + r2 = 14 +(r1 - r2 = 8) => 2r1 = 22

=>r1= 11cm

r1+r2=14 -(r1-r2=8) => 2r2 = 6

=>r2 = 6/2 = 3cm