Question-26205

Question Number 26205 by ajfour last updated on 22/Dec/17

Commented by ajfour last updated on 22/Dec/17

F i n d t h e a r e a o f t h e u n c o l o u r e d

r e g i o n i n s i d e t r i a n g l e i n t e r m s

o f r a d i i a, a n d b o f t h e l o w e r

c i r c l e s .

Answered by mrW1 last updated on 22/Dec/17

O n e s i d e o f t h e t r i a n g l e i s 2 (a + b) .

T h e o t h e r t w o s i d e s a r e d a n d t h e

r a d i u s o f t h e b i g c i r c l e i s c .

d = b + c .

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

b^{2} + 2 b c + c^{2} = a^{2} + b^{2} + 2 a b + a^{2} + 2 a c + c^{2}

(b - a) c = a (b + a)

\Rightarrow c = \frac{(b + a) a}{b - a} > 0

\Rightarrow b > a

a n g l e o f t r i a n g l e a t c e n t e r o f c i r c l e c :

\tan \frac{θ}{2} = \frac{b + a}{a + c} = \frac{b + a}{a + \frac{(b + a) a}{b - a}} = \frac{b^{2} - a^{2}}{2 a b}

\Rightarrow θ = {2 \tan}^{- 1} \frac{b^{2} - a^{2}}{2 a b} = π - 2 φ

a n g l e o f t r i a n g l e a t c e n t e r o f c i r c l e b :

\tan φ = \frac{a + c}{b + a} = \frac{2 a b}{b^{2} - a^{2}}

\Rightarrow φ = \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}

a r e a o f t r i a n g l e :

A_{1} = \frac{2 (a + b) (c + a)}{2} = (a + b) (\frac{b + a}{b - a} + 1) a = \frac{2 a b (a + b)}{b - a}

a r e a o f p a r t o f t r i a n g l e a :

A_{2} = \frac{π a^{2}}{2}

a r e a o f p a r t s o f t r i a n g l e b :

A_{3} = \frac{2 φ}{2 π} \times π b^{2} = φ b^{2} = b^{2} \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}

a r e a o f p a r t o f t r i a n g l e c :

A_{4} = \frac{θ}{2 π} \times π c^{2} = \frac{π - 2 φ}{2} \times c^{2}

= (\frac{π}{2} - \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}) \times \frac{a^{2} {(a + b)}^{2}}{{(b - a)}^{2}}

a r e a o f t h e r e s t :

A = A_{1} - (A_{2} + A_{3} + A_{4})

= \frac{2 a b (a + b)}{b - a} - \frac{π a^{2}}{2} - b^{2} \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}} - (\frac{π}{2} - \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}) \times \frac{a^{2} {(a + b)}^{2}}{{(b - a)}^{2}}

= \frac{2 a b (a + b)}{b - a} - \frac{π a^{2}}{2} [1 + \frac{{(a + b)}^{2}}{{(b - a)}^{2}}] - [b^{2} - \frac{a^{2} {(a + b)}^{2}}{{(b - a)}^{2}}] \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}

= \frac{2 a b (a + b)}{b - a} - π a^{2} [\frac{a^{2} + b^{2}}{{(b - a)}^{2}}] - [\frac{(a^{2} + b^{2}) (b^{2} - 2 a b - a^{2})}{{(b - a)}^{2}}] \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}

\Rightarrow A = \frac{2 a b (a + b)}{b - a} - \frac{a^{2} + b^{2}}{{(b - a)}^{2}} \times [π a^{2} + (b^{2} - a^{2} - 2 a b) \tan^{- 1} \frac{2 a b}{b^{2} - a^{2}}]

Commented by ajfour last updated on 22/Dec/17

G r e a t S i r, M a g n i f i c i e n t!

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