With the vertices of $△ A B C$ as centres, three circles are described, each touching the other two externally. If the sides of the triangle are $9 c m, 7 c m$ and $6 c m$ , The radii of the circles with centres A, B & C are

2 c m, 5 c m a n d 4 c m r e s p e c t i v e l y .

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4 c m, 5 c m a n d 2 c m r e s p e c t i v e l y .

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5 c m, 2 c m a n d 4 c m r e s p e c t i v e l y .

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2 c m, 4 c m a n d 5 c m r e s p e c t i v e l y .

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Solution

The correct option is B $4 c m, 5 c m a n d 2 c m r e s p e c t i v e l y .$
$G i v e n - A, B & C a r e t h e c e n t r e s o f t h r e e c i r c l e s t o u c h i n g e a c h o t h e e x t e r n a l l y .$

$A B = 9 c m, B C = 7 c m & C A = 6 c m .$

$T o f i n d o u t -$

$T h e r e s p e c t i v e r a d i i o f t h e c i r c l e s .$

$S o l u t i o n -$

$L e t x, y & z b e t h e r a d i i o f t h e c i r c l e s w i t h c e n t r e s A, B & C r e s p e c t i v e l y .$

$W e k n o w t h a t t h e d i s t a n c e b e t w e e n t h e c e n t r e s o f t w o c i r c l e s, w h o t o u c h e x t e r n a l l y, i s t h e s u m o f t h e i r r a d i i .$

$∴ x + y = 9 c m . . . . . . (i), y + z = 7 c m . . . . . . . (i i) a n d z + x = 6 c m . . . . . (i i i)$

$A d d i n g i, i i a n d i i i w e g e t x + y + z = \frac{9 + 7 + 6}{2} c m = 11 c m . . . . . . (i v) .$

$S u b t r a c t i n g i f r o m i v w e g e t z = (11 - 9) c m = 2 c m,$

$s u b t r a c t i n g i i f r o m i v w e g e t x = (11 - 7) c m = 4 c m a n d$

$s u b t r a c t i n g i i i f r o m i v w e g e t y = (11 - 6) c m = 5 c m .$

$S o t h e r a d i i o f t h e c i r c l e s w i t h c e n t r e A, B & C a r e 4 c m, 5 c m a n d 2 c m r e s p e c t i v e l y . A n s - O p t i o n B .$

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