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Question Number 15170 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Jun/17

$A B C D, i s a s q u a r e w i t h a r e a = 1$ $e a c h a c u t e a n g l e s s u c h t h a t : ∡ {A D A}^{'}$ $a r e e q u a i l t o : 15^{∙}$ $1) s h o w t h a t : t h e w h i t e s h a p e i s a s q u a r e .$ $2) f i n d {i t}^{'} s a r e a .$ $3) s h o w t h a t : {D D}^{'} i s t h e b i s e c t o f ∡ {A D A}^{'} .$ $4) f i n d p a r t o f A {\overset{Δ}{D A}}^{'} s a r e a t h a t l o c a t e d$ $o n t h e r i g h t s i d e o f l i n e {A B}^{'} .$
Commented by ajfour last updated on 08/Jun/17

$i t s h o u l d b e g i v e n, t h a t A^{'} D = A D . . .$
	Answered by mrW1 last updated on 08/Jun/17

	$side length of the white square :$ $p = 1 \times \cos 15 ° - 1 \times \sin 15 °$ $= \sqrt{\frac{1 + \cos 30 °}{2}} - \sqrt{\frac{1 - \cos 30 °}{2}}$ $= \frac{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}{2}$ $area of white square :$ $p^{2} = {(\frac{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}{2})}^{2} = \frac{1}{2}$ $∠ A^{'} {AB}^{'} = \frac{180 - 15}{2} - (90 - 15) = 7.5 °$ $\Rightarrow A^{'} A is bisector of ∠ {BAB}^{'}$ $let M = intersection of {DA}^{'} and {AB}^{'}$ $A^{'} M = {DA}^{'} - DM = 1 - \cos 15 °$ $area of Δ {AMA}^{'}$ $= \frac{1}{2} A^{'} M \times AM = \frac{1}{2} \times (1 - \cos 15 °) \times \sin 15 °$ $= \frac{1}{2} \times (\sin 15 ° - \frac{\sin 30 °}{2})$ $= \frac{1}{2} \times (\frac{\sqrt{2 - \sqrt{3}}}{2} - \frac{1}{4})$ $= \frac{2 \sqrt{2 - \sqrt{3}} - 1}{8} \approx 0.0044$
	Commented by mrW1 last updated on 08/Jun/17

	$you are right, thanks . it is fixed now .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

	$t h a n k y o u m y m a s t e r . b u t w a y o u r$ $a n s w e r s a r e d i f f e r e n t ?$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

	$d e a r m a s t e r m r W 1! i t h i n k r e l a t i o n :$ $\frac{A^{'} M}{A M} = \frac{A M}{D M}, i s n o t c o r r e c t .$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

	$A D = 1$ $M, i s t h e i n t e r s e c t o f : {D A}^{'}, {A B}^{'} .$ $s i n 15 = \frac{A M}{1} \Rightarrow A M = s i n 15 = . 2588$ $c o s 15 = \frac{D M}{1} \Rightarrow D M = c o s 15 = . 9659$ $t g \frac{15}{2} = \frac{{M A}^{'}}{M A} \Rightarrow {M A}^{'} = M A . t g 7.5$ $t g 7.5 = \frac{s i n 15}{1 + c o s 15} = 0.1317$ $\Rightarrow {M A}^{'} = . 2588 \times . 1317 = . 0341$ $S = \frac{1}{2} . {M A}^{'} . M A = \frac{1}{2} \times 0.0341 \times . 2588 = . 00441$
	Commented by mrW1 last updated on 08/Jun/17

	$I think here is error$ $\Rightarrow {M A}^{'} = \frac{\sqrt{6} - \sqrt{2}}{4} . (1 - \frac{{(\sqrt{3} - \sqrt{2} + 1)}^{2}}{2}) =$ $\neq \frac{\sqrt{2}}{4} (2 \sqrt{2} - \sqrt{3} + 1) (error)$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

	$y e s . i t i s n o w c o r r e c t e d . t h a n k s .$
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