Question-181055

Question Number 181055 by Acem last updated on 21/Nov/22

Commented by Acem last updated on 21/Nov/22

* D A B C

P l e a s e, t h e c o m m e n t s s e c t i o n i s o n l y

f o r

i n q u i r i e s a n d c l a r i f i c a t i o n s

Commented by som(math1967) last updated on 21/Nov/22

Commented by som(math1967) last updated on 21/Nov/22

∠ D O A = 2 ∠ D L A

s o O i s c e n t r e o f c i r c u m c i r c l e o f

△ L D A O D = O A = \frac{10}{\sqrt{2}} = 5 \sqrt{2}

C B ∥ D A ∴ △ O D A \sim △ O B C

\frac{O B}{O D} = \frac{B C}{D A} = \frac{4}{10}

\Rightarrow O B = \frac{2}{5} \times 5 \sqrt{2} = 2 \sqrt{2}

a r o f D A B C = △ A O D + △ A O B

+ △ B O C + △ D O C

= \frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2} + \frac{1}{2} \times 2 \sqrt{2} \times 5 \sqrt{2}

+ \frac{1}{2} \times 2 \sqrt{2} \times 2 \sqrt{2} + \frac{1}{2} \times 2 \sqrt{2} \times 5 \sqrt{2}

= 25 + 10 + 4 + 10 = 49 s q u n i t

Commented by mr W last updated on 21/Nov/22

A c e m s i r : w i t h a n g l e 45 °, t h e t r a n g l e

a n d t h e t r a p a z o i d a r e s y m m e t r i c,

a n d t h e s o l u t i o n i s o b v i o u s . w e c a n

g e t t h e r e s u l t d i r e c t l y a s s o m s i r d i d .

i f t h e a n g l e i s 60 °, w e {c a n}^{'} t g e t t h e

r e s u l t o b v i o u s l y a s w i t h 45 ° .

w i t h 45 °, t h e d i a g o n a l s o f t h e t r a p a z o i d

a r e e q u a l a n d e q u a l t o (5 + 2) \sqrt{2} = 7 \sqrt{2} .

t h e a r e a o f t h e t r a p a z o i d i s t h e n

\frac{7 \sqrt{2} \times 7 \sqrt{2}}{2} = 49 .

w i t h a n g l e 60 °, w e {c a n}^{'} t g e t i t s o

o b v i o u s l y .

Commented by Acem last updated on 21/Nov/22

W h a t i f D o r A w a s o u t o f t h e c i r c l e ?

I n o t h e r w o r d, w e c a n f o r m i n f i n i t e n u m b e r o f

t r a p e z o i d s w i t h 2 p a r a l l e l b a s e s .

A n d p l e a s e c o p y a l l y o u r s o l u t i o n s t o b u f f e r f r o m

t h e m e n u ⋮ a n d p a s t e c o p i e d b u f f e r a s a n s w e r

p a r t, t h e n p l e a s e r e m o v e t h e c o m m e n t s .

T h a n k s f r i e n d!

Commented by mr W last updated on 21/Nov/22

t h e a n g l e 45 ° m a k e s t h e t h i n g t o a

s p e c i a l c a s e w h i c h i s t o o o b v i o u s .

{i t}^{'} s m o r e i n t e r e s t i n g w h e n t h e a n g l e

i s n o t 45 °, b u t e . g . 60 ° .

Commented by mr W last updated on 21/Nov/22

t h e s o l u t i o n a b o v e f r o m s o m s i r i s

w o n d e r f u l!

Commented by Acem last updated on 21/Nov/22

N o p r o b l e m S i r S o m

Commented by Acem last updated on 21/Nov/22

W i t h a n a n g l e 45 i s m o r e i n t e r e s t i n g t h a n 60

Commented by som(math1967) last updated on 21/Nov/22

s o r r y c a n n o t l o a d a s a n s w e r

Commented by MJS_new last updated on 21/Nov/22

they are n o t symmetric in my opinion . the

angle in the center is marked as 90 °

Commented by mr W last updated on 21/Nov/22

y e s . n o w i s e e . i t {m u s t n}^{'} t b e s y m m e t r i c

w i t h 45 ° .

Commented by MJS_new last updated on 21/Nov/22

the difference is not visible in the drawing .

with a base length 10 and an angle of 45 °

between the bottom

l line and the diagonals we {don}^{'} t get length

4 for the shorter horizontal line but

10 (\sqrt{2} - 1) \approx 4.14

Commented by mr W last updated on 21/Nov/22

y e s s i r! i s a w t h i s c o n t r a d i c t i o n v e r y

l a t e . i t h i n k, i t i s s y m m e t r i c o n l y

w h e n θ = 46.397 ° .

Answered by MJS_new last updated on 21/Nov/22

the angle ∡ D O A = 90 ° (or else use a different

symbol in your drawing)

let L = (\begin{matrix} 0 \\ 0 \end{matrix}) \land C = (\begin{matrix} p \\ 0 \end{matrix}) \land B = (\begin{matrix} q \\ q \end{matrix})

similar triangles \Rightarrow

D = (\begin{matrix} 5 p / 2 \\ 0 \end{matrix}) \land A = (\begin{matrix} 5 q / 2 \\ 5 q / 2 \end{matrix})

we need ∣ B C ∣^{2} = 16 (\Leftrightarrow ∣ A D ∣^{2} = 100) \Rightarrow

p^{2} - 2 p q + 2 q^{2} = 16 (1)

A C : y = - \frac{5 q}{2 p - 5 q} x + \frac{5 p q}{2 p + 5 q}

B D : y = - \frac{2 q}{5 p - 2 q} x + \frac{5 p q}{5 p - 2 q}

A C ⊥ B D \Rightarrow

- \frac{5 q}{2 p - 5 q} = \frac{5 p - 2 q}{2 q} \Leftrightarrow

p^{2} - \frac{29}{10} p q + 2 q^{2} = 0 (2)

(1) - (2) \Rightarrow q = \frac{160}{9 p}

insert in (1) or (2) \Rightarrow

p^{4} - \frac{464}{9} p^{2} + \frac{51200}{81} = 0

p > 0 \Rightarrow p_{1} = \frac{2 \sqrt{2 (29 - \sqrt{41})}}{3} \land p_{2} = \frac{2 \sqrt{2 (29 + \sqrt{41})}}{3}

due to symmetry ∣ {L C}_{p_{1}} ∣=∣ {L B}_{p_{2}} ∣ I take p_{2} \Rightarrow

p = \frac{2 \sqrt{2 (29 + \sqrt{41})}}{3} \Rightarrow q = \frac{2 \sqrt{29 - \sqrt{41}}}{3} ★

\Rightarrow

area triangle L D A = \frac{500}{9}

area triangle L C B = \frac{80}{9}

area trapezoid = \frac{140}{3}

Commented by Acem last updated on 21/Nov/22

V e r y w e l l w o r k S i r!

Y o u c a n c h e c k t h e o t h e r m e t h o d b e l l o w

T h a n k s!

Answered by Acem last updated on 21/Nov/22

Commented by Acem last updated on 21/Nov/22

W e h a v e a d = \frac{40}{21} & 4 d^{2} + 4 a^{2} = 16

b y c o m m o n s o l u t i o n :

441 d^{4} - 1764 d^{2} + 1600 = 0

w e w i l l f i n d 2 s o l u t i o n s

a_{1} \approx 1.17905 d_{1} \approx 1.6155

a_{2} \approx 1.6155 d_{2} \approx 1.17905

F o r e x a m p l e w e t a k e t h e 2 n d o n e :

A M = 5 d \approx 5.895

D M = 5 a \approx 8.077 t h e y a r e n o t e q u a l

T h e c i r c l e o f M a s c e n t e r w i t h r = A M,

t h e v e r t i c e s D a n d F w i l l b e b o t h o u t o f i t .

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