Question

$can \: ya \: all \: please \: help$
Two circles touch internally. The sum of their areas is 116 π cm² and the distance between their centres is 6 cm. Find the radii of the circles.

Answer 1

Answer:

10 cm, 4 cm

Step-by-step explanation:

Let R and r be the radii of the circles having centres at O and O' respectively

(i)

Sum of their areas is 116 π.

⇒ πR² + πr² = 116π

⇒ π(R² + r²) = 116π

⇒ R² + r² = 116

(ii)

Distance between their centres is 6 cm.

⇒ OO' = 6 cm.

⇒ (R - r) = 6 cm

(iii)

We know that (R + r)² + (R - r)² = 2(R² + r²)

⇒ (R + r)² + (6)² = 2(116)

⇒ (R + r)² = 232 - 36

⇒ (R + r)² = 196

⇒ R + r = 14

On solving (ii) & (iii), we get

R - r = 6

R + r = 14

--------------

2R = 20

R = 10 cm.

Substitute R = 10 in (ii), we get

R - r = 6

10 - r = 6

r = 4 cm.

Hence, radii of the two circles are 10 cm and 4 cm respectively.

Hope it helps!

Answer 2

✨✨hey mate here the solution ✨✨✨


Given:

Sum of areas of two circles = 116*(pi) cm^2

Distance between the centers of two circles = 6 cm

Since the two circles are touching internally, the distance between the centers of the two circles =

(Radius of bigger circle) - (Radius of smaller circle)

Let r = radius of bigger circle

Then radius of small circle = r - 6

Area of big circle = (pi)*r^2

and area of small circle = 116*(pi)  ... (given)

= (pi)*(r - 6)^2 = (pi/4)*(r^2 - 12r + 36)

Sum of area of two circles = (pi)*r^2 + [(pi/4)*(r^2 - 12r + 36)]

= (pi)*(2r^2 - 12r + 36)

Dividing both side of equation by 2*(pi) we get:

58 = r^2 - 6r + 18

Transferring all terms of the above equation on one side we get:

r^2 - 6r - 40 = 0

by factorising left hand side of the equation we get:

(r - 10)(r + 4) = 0

Therefor r = 10, or r = -4

As radius cannot be negative, the radius of bigger circle =  10 cm.

And the radius of small er circle = 10 - 6 = 4 cm.

✨✨thanks