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Question Number 26853 by ajfour last updated on 30/Dec/17

Commented by ajfour last updated on 30/Dec/17

$F i n d a r e a o f t h e b l u e r e g i o n .$
	Answered by mrW1 last updated on 30/Dec/17

	Commented by mrW1 last updated on 31/Dec/17

	$D (1, 1)$ $x_{C} = 1$ $y_{C} = \frac{1}{a}$ $x_{B} = \frac{a}{2}$ $y_{B} = \frac{1}{2}$ $x_{A} = y_{A}$ $\frac{x_{A}}{1} = \frac{a - x_{A}}{a}$ $\Rightarrow x_{A} = \frac{a}{1 + a} = y_{A}$ $A_{A B C D} = \frac{1}{2} (1 - \frac{a}{1 + a}) (1 + \frac{a}{1 + a}) - \frac{1}{2} (\frac{a}{2} - \frac{a}{1 + a}) (\frac{1}{2} + \frac{a}{1 + a}) - \frac{1}{2} (1 - \frac{a}{2}) (\frac{1}{2} + \frac{1}{a})$ $A_{H e x a g o n} = 2 \times A_{A B C D} = (1 - \frac{a}{1 + a}) (1 + \frac{a}{1 + a}) - (\frac{a}{2} - \frac{a}{1 + a}) (\frac{1}{2} + \frac{a}{1 + a}) - (1 - \frac{a}{2}) (\frac{1}{2} + \frac{1}{a})$ $= \frac{1 + 2 a}{{(1 + a)}^{2}} - \frac{a (a - 1) (1 + 3 a)}{4 {(1 + a)}^{2}} - \frac{(2 - a) (2 + a)}{4 a}$ $= \frac{4 (1 + 2 a) - a (a - 1) (1 + 3 a)}{4 {(1 + a)}^{2}} - \frac{1}{a} + \frac{a}{4}$ $= \frac{4 + 3 a^{3} + 2 a^{2} + 9 a}{4 {(1 + a)}^{2}} - \frac{1}{a} + \frac{a}{4}$ $. . . . . .$
	Commented by ajfour last updated on 31/Dec/17

	$t h a n k y o u S i r, q u i t e e n l i g h t e n i n g$ $f o r m e .$
	Commented by mrW1 last updated on 31/Dec/17

	Commented by ajfour last updated on 31/Dec/17

	$i n l a s t l i n e y o u s h o u l d h a v e - 3 a^{3}$ $i n s t e a d o f 3 a^{3} .$ $F r o m y o u r 4^{t h} l a s t l i n e S i r,$ $i f i e x p a n d :$ $A_{h e x a} = 1 - {(\frac{a}{1 + a})}^{2} - \frac{a}{4} - \frac{a^{2}}{2 (1 + a)}$ $+ \frac{a}{2 (1 + a)} + {(\frac{a}{1 + a})}^{2} - \frac{1}{2} - \frac{1}{a}$ $+ \frac{a}{4} + \frac{1}{2}$ $A_{h e x a} = 1 - \frac{a (a - 1)}{2 (1 + a)} - \frac{1}{a} .$ $s h o r t e n o u g h .$
	Commented by ajfour last updated on 31/Dec/17

	$A_{h e x a g o n} = 2 [A r (△ O C D) - A r (△ O A B)$ $= 2 [\frac{1}{2} - \frac{1}{2 a} - \frac{1}{2} \| \begin{matrix} \frac{a}{1 + a} & \frac{a}{1 + a} & 1 \\ \frac{a}{2} & \frac{1}{2} & 1 \\ 0 & 0 & 1 \end{matrix} \|]$ $= 1 - \frac{1}{a} - \frac{a (a - 1)}{2 (1 + a)} .$
	Commented by mrW1 last updated on 31/Dec/17

	$G r e a T!$
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