Question-79609

Question Number 79609 by mr W last updated on 26/Jan/20

Commented by mr W last updated on 26/Jan/20

i f t h e l a r g e s q u a r e h a s t h e s i d e l e n g t h

10, f i n d t h e r a d i u s r o f t h e s m a l l e s t

c i r c l e .

Answered by ajfour last updated on 26/Jan/20

l e t b i g s q u a r e s i d e = 10 = 2 a

s i d e o f s m a l l e r o n e = 2 b

c i r c l e r a d i u s = r

4 a^{2} = 4 b^{2} + {(a + b)}^{2}

\Rightarrow 3 a^{2} = 5 b^{2} + 2 a b

λ = \frac{a}{b} \Rightarrow 3 λ^{2} - 2 λ - 5 = 0

λ = \frac{1}{3} + \sqrt{\frac{1}{9} + \frac{5}{3}} = \frac{5}{3}

{(2 a - r)}^{2} = a^{2} + {(2 b + r)}^{2}

\Rightarrow

3 a^{2} - 4 a r = 4 b^{2} + 4 b r

r = \frac{3 a^{2} - 4 b^{2}}{4 (a + b)} = \frac{a (3 \times \frac{25}{9} - 4)}{4 (\frac{5}{3} + 1)}

= \frac{13 a}{32} = \frac{65}{32} .

Commented by ajfour last updated on 26/Jan/20

Commented by mr W last updated on 26/Jan/20

t h a n k s f o r t h e s o l u t i o n s i r!

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