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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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