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Question Number 154475 by mr W last updated on 18/Sep/21

Commented by mr W last updated on 18/Sep/21

$f i n d t h e a r e a o f t h e b i g t r i a n g l e w h o s e$ $s i d e s h a v e t h e d i s t a n c e s d_{1}, d_{2}, d_{3} t o$ $t h e s i d e s a, b, c o f t h e s m a l l t r i a n g l e .$
Commented by talminator2856791 last updated on 18/Sep/21

$are the lines parallel ?$
Commented by mr W last updated on 18/Sep/21

$y e s, i t i s s a i d .$
Commented by Tawa11 last updated on 21/Sep/21

$great sir$
	Answered by mr W last updated on 19/Sep/21

	$l e t Δ = a r e a o f t r i a n g l e A B C$ $Δ = \sqrt{s (s - a) (s - b) (s - c)}$ $w i t h s = \frac{a + b + c}{2}$ $s a y t h e i n c e n t e r o f t r i a n g l e A B C i s E$ $a n d t h e r a d i u s o f i n c i r c l e i s r .$ $r = \frac{Δ}{s} = \frac{2 Δ}{a + b + c}$ $s a y t h e d i s t a n c e s f r o m E t o t h e s i d e s$ $o f t h e t r i a n g l e A^{'} B^{'} C^{'} a r e h_{1}, h_{2}, h_{3}$ $r e s p e c t i v e l y, w e h a v e$ $h_{1} = r + d_{1}$ $h_{2} = r + d_{2}$ $h_{3} = r + d_{3}$
	Commented by mr W last updated on 19/Sep/21

	Commented by mr W last updated on 20/Sep/21

	$i t i s o b v i o u s t h a t b o t h t r i a n g l e s a r e$ $s i m i l a r, s i n c e t h e i r s i d e s a r e p a r a l l e l$ $t o e a c h o t h e r .$ $s a y t h e s i d e l e n g t h e s o f t h e b i g$ $t r i a n g l e a r e a^{'}, b^{'}, c^{'} w i t h$ $a^{'} = k a$ $b^{'} = k b$ $c^{'} = k c$ $w h e r e k i s t h e m a g n i f i c a t i o n f a c t o r .$ $s a y t h e a r e a o f t h e b i g t r i a n g l e A^{'} B^{'} C^{'}$ $i s Δ^{'} . t h e n w e h a v e Δ^{'} = k^{2} Δ .$ $o n t h e o t h e r s i d e w e h a v e$ $Δ^{'} = \frac{a^{'} \times h_{1}}{2} + \frac{b^{'} \times h_{2}}{2} + \frac{c^{'} \times h_{3}}{2}$ $Δ^{'} = \frac{k a \times (r + d_{1})}{2} + \frac{k b \times (r + d_{2})}{2} + \frac{k c \times (r + d_{3})}{2}$ $Δ^{'} = \frac{k}{2} [(a + b + c) r + {a d}_{1} + {b d}_{2} + {c d}_{3}]$ $Δ^{'} = \frac{k}{2} [(a + b + c) \frac{2 Δ}{(a + b + c)} + {a d}_{1} + {b d}_{2} + {c d}_{3}]$ $Δ^{'} = \frac{k}{2} (2 Δ + {a d}_{1} + {b d}_{2} + {c d}_{3})$ $s i n c e Δ^{'} = k^{2} Δ,$ $\frac{k}{2} (2 Δ + {a d}_{1} + {b d}_{2} + {c d}_{3}) = k^{2} Δ$ $\Rightarrow k = 1 + \frac{{a d}_{1} + {b d}_{2} + {c d}_{3}}{2 Δ}$ $t h e r e f o r e t h e a r e a o f b i g t r i a n g l e i s$ $Δ^{'} = {(1 + \frac{{a d}_{1} + {b d}_{2} + {c d}_{3}}{2 Δ})}^{2} Δ$
	Commented by Rasheed.Sindhi last updated on 20/Sep/21

	$W onderful S ir!$
	Commented by mr W last updated on 20/Sep/21

	$t h a n k s f o r r e v i e w i n g s i r!$
	Answered by Rasheed.Sindhi last updated on 19/Sep/21

	Commented by Rasheed.Sindhi last updated on 19/Sep/21

	$△ ABC \sim △ A^{'} B^{'} C^{'}$ $△ ABC h a s b e e n m o v e d s o t h a t$ $A c o i n c i d e s A^{'}$ $∠ BAC c o i n s i d e s ∠ B^{'} A^{'} C^{'}$ $d 2 = d 3 = 0 i n t h i s c a s e .$ $C o n t i n u e$
	Commented by mr W last updated on 19/Sep/21

	$\frac{{a h}_{a}}{2} = Δ$ $h_{a} = \frac{2 Δ}{a}$ $Δ^{'} = {(\frac{h_{a} + d_{1}}{h_{a}})}^{2} Δ = {(1 + \frac{d_{1}}{\frac{2 Δ}{a}})}^{2} Δ = {(1 + \frac{{a d}_{1}}{2 Δ})}^{2} Δ$ $u s i n g m y f o r m u l a w i t h d_{2} = d_{3} = 0$ $Δ^{'} = {(1 + \frac{{a d}_{1}}{2 Δ})}^{2} Δ ✓$
	Commented by mr W last updated on 19/Sep/21

	$i h a v e r e v i s e d m y s o l u t i o n a b o v e .$ $p l e a s e c r i t i c a l r e v i e w! t h a n k s!$
	Commented by Rasheed.Sindhi last updated on 19/Sep/21

	$T h a n k s t o h e l p f o r c o m p l e t i n g m y$ $a p p r o a c h a l s o .$
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