Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 182006 by mr W last updated on 03/Dec/22

Commented by mr W last updated on 03/Dec/22

$f i n d t h e l a r g e s t a r e a o f a n i n s c r i b e d$ $r i g h t - a n g l e t r i a n g l e i n a g i v e n a c u t e$ $t r i a n g l e w i t h s i d e s a, b, c . (a ⩾ b ⩾ c)$
Commented by Acem last updated on 03/Dec/22

$I n t e r e s t i n g o n e!$
	Answered by mr W last updated on 03/Dec/22

	Commented by mr W last updated on 03/Dec/22

	$A_{m a x} = \frac{a^{2} \sin γ \cos γ}{2} = \frac{a^{2} \sin 2 γ}{4}$
	Commented by Acem last updated on 03/Dec/22

	$T h e 2 n d {o n e}_{b e l o w} = \frac{a^{2} \sin β \cos β}{2} = \frac{a^{2} \sin 2 β}{4}$ $β > γ t h e n t h e 2 n d t r i a n g l e i s l a r g e r t h a n t h e 1 s t$
	Commented by mr W last updated on 03/Dec/22

	${d o n}^{'} t y o u s e e t h a t β > γ {d o e s n}^{'} t$ $a u t o m a t i c a l l y m e a n t h a t t h e s e c o n d$ $t r i a n g l e i s l a r g e r t h a n t h e f i r s t o n e ?$ $l o o k a t t h e d i a g r a m b e l o w .$
	Commented by mr W last updated on 03/Dec/22

	Commented by mr W last updated on 03/Dec/22

	Commented by mr W last updated on 03/Dec/22

	$t h i s p i c t u r e s h o w s t h a t t h e f i r s t$ $t r i a n g l e (= \frac{a^{2} \sin 2 γ}{4}) i s a l w a y s l a r g e r$ $t h a n t h e s e c o n d o n e (= \frac{a^{2} \sin 2 β}{4}) .$
	Commented by Acem last updated on 03/Dec/22

	Commented by Acem last updated on 03/Dec/22

	$W e m a y s a y h e r e t h a t △ M A C > B A N$ ${I t}^{'} s o k B u t, {A r e a}_{M A C} = \frac{1}{2} b ∣ A M ∣= \frac{1}{2} b^{2} \tan γ$ $S o h o w c a n w e j u d g e ? c a u s e a l l y o u r p o s t s m a d e$ $m e t h i n k t h a t t h e m a x i m u m a r e a i s w h e n t h e$ $r i g h t a n g l e b e o n t h e 2 n d s i d e ‘ ‘ m i d d l e {l e n g t h}^{″} &$ $t h e h y p o t e n u s e i s w h o l e o f t h e s i d e ‘ ‘ a^{″} w i t h$ $a f o r m u l a \frac{a^{2} \sin 2 γ}{4} \overset{?}{∣} \frac{1}{2} b^{2} \tan γ$ $N o t e : I t h o u g h t t h r o u g h d i a g r a m 1 t h a t y o u s a y$ $t h a t {t h e r e}^{'} s 2 t r i a n g l e s a s m a x i m u m$
	Commented by Acem last updated on 04/Dec/22
	$I back to your question, you mentioned that it is an acute triangle, well can now say? : For a≥ b ≥ c then: maximum { ((in an cute tri. ((a^2 sin 2γ)/4))),((in obtuse tri. ((b^2 tan γ)/2) _(You agree?) )) :}$
	$I b a c k t o y o u r q u e s t i o n, y o u m e n t i o n e d t h a t i t i s$ $a n a c u t e t r i a n g l e, w e l l c a n n o w s a y ? :$ $F o r a ⩾ b ⩾ c t h e n :$ $m a x i m u m {\begin{cases} i n a n c u t e t r i . \frac{a^{2} \sin 2 γ}{4} \\ i n o b t u s e t r i . \frac{b^{2} \tan γ}{2}_{Y o u a g r e e ?} \end{cases}$
	Commented by mr W last updated on 04/Dec/22

	$a g r e e .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum