$P Q R$ is a right-angled triangle with $P Q = 3 c m$ and $Q R = 4 c m$ . A circle which touches all the sides of the triangle is inscribed in the triangle. The radius of the circle is

2cm

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1cm

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5cm

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3cm

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Solution

The correct option is B 1cm
$G i v e n - P Q R i s a r i g h t t r i a n g l e w i t h ∠ P Q R = 1 r i g h t a n g l e .$

$Δ P Q R i n s c r i b e s a c i r c l e w i t h c e n t r e a s O .$

$R Q t o u c h e s t h e i n c i r c l e a t D,$

$R P t o u c h e s t h e i n c i r c l e a t E a n d$

$P Q t o u c h e s t h e i n c i r c l e a t F .$

$Q R = 4 c m & P Q = 3 c m . T o f i n d o u t - t h e r a d i u s o f t h e i n c i r c l e = (O D, O E, O F) = x = ?$

$S o l u t i o n -$

$G i v e n t h a t P Q R i s a r i g h t t r i a n g l e w i t h ∠ P Q R = 1 r i g h t a n g l e$

$So,\quad applying\quad Pythagoras\quad theorem,\quad we\quad get

$P Q = \sqrt{{R P}^{2} + {P Q}^{2}} = \sqrt{4^{2} {+ 3}^{2}} c m = 5 c m .$

$N o w R D = R E = p, Q D = Q F = q a n d P F = P E = z . s i n c e t h e l e n g t h s o f t h e t a n g e n t s, f r o m a p o i n t t o a c i r c l e, a r e e q u a l .$

$∴ (p + q) + (q + r) + (r + p) = R Q + P Q + R P = (4 + 3 + 5) c m = 12 c m$

$p + q + r = \frac{12}{2} c m = 6 c m . . . . . . . . . . (i) A l s o R Q = p + q = 4 c m, P Q = r + q = 3 c m & R P = r + p = 5 c m . (i) - R P = q = (6 - 5) c m = 1 c m . . . . . . . . . . . . (i i) .$

$N o w O D ⊥ A B & O F ⊥ B C . s i n c e t h e r a d i u s t h r o u g h t h e p o i n t o f c o n t a c t o f a t a n g e n t t o a c i r c l e i s p e r p e n d i c u l a r t o t h e t a n g e n t . ∴ ∠ Q D O = ∠ B F O = 90^{o} . ∴ I n B F O D a l l t h e a n g l e s a r e e q u a l a n d B D = B F . S o B F O D i s a s q u a r e . i . e x = B D = q = 1 c m . ∴ T h e r a d i u s o f t h e i n c i r c l e = x = 1 c m . A n s - O p t i o n B .$

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