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Construction of a Triangle with ASA Criterion

An isosceles ...

Question

An isosceles $△$ ABC is inscribed in a circle. If $A B = A C = 12$ $\sqrt{5} c m$ and $B C = 24 c m$ , then the radius of the circle is:

A

12 c m

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B

24 c m

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C

15 c m

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D

18 c m

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Solution

The correct option is C $15 c m$
$G i v e n - O i s t h e c e n t r e o f a c i r c l e w h i c h h a s t w o c h o r d s A B = A C = 12 \sqrt{5} c m . T h e l e n g t h o f t h e c h o r d B C = 24 c m . T o f i n d o u t - T h e r a d i u s o f t h e c i r c l e = ? S o l u t i o n - I n Δ A B C A B = A C = 12 \sqrt{5} c m . . ∴ Δ A B C i s a n i s o s c e l e s o n e w i t h B C a s b a s e . C o n s t r u c t i o n - T h e b i s e c t o r o f ∠ B A C i s d r a w n a n d i t i n t e r s e c t s B C a t D . N o w t h e b i s e c t o r o f t h e a n g l e b e t w e e n t h e e q u a l s i d e s o f a n i s o s c e l e s t r i a n g l e i s t h e p e r p e n d i c u l a r b i s e c t o r o f t h e b a s e . ∴ A D ⊥ B C ⟹ B D = D C = \frac{1}{2} B C = \frac{1}{2} \times 24 c m = 12 c m . . . . . . . (i) . A l s o ∠ A D B = ∠ A D C = 90^{9} . B u t w e k n o w t h a t t h e p e r p e n d i c u l a r, d r o p p e d f r o m t h e c e n t e r o f a c i r c l e t o i t s a n y c h o r d b i s e c t s t h e l a t t e r . ∴ O l i e s o n A D . i . e ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ A O D i s a s t r a i g h t l i n e . W e j o i n O C . ∴ O C i s a r a d i u s o f t h e g i v e n c i r c l e . S o O C = O A = r (s a y) N o w Δ A D C i s a r i g h t o n e w i t h A C a s h y p o t e n u s e . S o, b y P y t h a g o r a s t h e o r e m, w e h a v e A D = \sqrt{{A C}^{2} - {D C}^{2}} = \sqrt{{(12 \sqrt{5})}^{2} - 12^{2}} c m = 24 c m . ∴ O D = A D - r = 24 - r . . . . . . . . (i i) N o w O D C i s a r i g h t o n e w i t h A C a s h y p o t e n u s e a s ∠ O D C = 90^{9} (f r o m i) . S o, b y P y t h a g o r a s t h e o r e m, w e h a v e {O D}^{2} + {D C}^{2} = {O C}^{2} ⟹ {(24 - r)}^{2} + 12^{2} = r^{2} ⟹ r = 15 c m . ∴ T h e r a d i u s o f t h e c i r c l e = 15 c m A n s - O p t i o n C .$

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△ A B C

¯ ¯¯¯¯¯¯ ¯ A C

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. The radius of the inscribed circle is:

Q. An isosceles

Δ A B C

is inscribed in a circle. If AB= AC =

12 \sqrt{5}

cm and BC = 24 cm. The radius of the circle is

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(a) $1$ cm (b) $2$ cm (c) $3$ cm (d) $4$ cm

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