Question-15318

Question Number 15318 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

g i v e n : A B C D, r e c t a n g l e . c o n n e c t

e a c h v e r t e x t o t h e m i d d l e p o i n t o f

o n e o t h e r s i d e s a s s h o w n .

1) s h o w t h a t M N P Q, i s a p a r a l l e l o g r a m

a n d i t s a r e a i s e q u a i l t o : \frac{1}{5} a r e a o f A B C D .

2) S_{S \overset{Δ}{P C}} = S_{B \overset{Δ}{N T}} = \frac{1}{20} . S_{A B C D} .

Commented by mrW1 last updated on 09/Jun/17

Commented by mrW1 last updated on 09/Jun/17

S_{1} = S_{2} = S_{3} = S_{4} = S_{5}

S_{1} + S_{2} + S_{3} + S_{4} + S_{5} = S_{ABCD}

\Rightarrow S_{1} = \frac{1}{5} S_{ABCD}

S_{Δ SPC} = \frac{1}{4} S_{4} = \frac{1}{4} \times \frac{1}{5} S_{ABCD} = \frac{1}{20} S_{ABCD}

S_{Δ BNT} = \frac{1}{4} S_{3} = \frac{1}{4} \times \frac{1}{5} S_{ABCD} = \frac{1}{20} S_{ABCD}

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

t h a n k y o u m a s t e r . i t i s s i m p l e a n d p e r f e c t .

Commented by mrW1 last updated on 09/Jun/17

what if not the middle point but the

1 / 3 - point is connected with the

vertex ?

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

i n c a s e o f : \frac{1}{3}, \frac{S_{A B C D}}{S_{M N P Q}} = 2.5, \frac{S_{A B C D}}{S_{B N T}} = 60 .

Answered by ajfour last updated on 10/Jun/17

Commented by ajfour last updated on 10/Jun/17

a l l / / g m s t h a t s e e m l i k e t h e

c e n t r a l o n e a r e a l i k e, w h i c h

i s n o t s o i n d i a g r a m, b u t c a n

b e j u d g e d s o .

Then 5 such / / gm areas =

A r e a o f r e c t a n g l e

T h e c o r n e r Δ^{'} s h a v e o n e - f o u r t h

o f t h e / / g m a r e a = o n e - t w e n t i e t h

o f t h e r e c t a n g l e a r e a .

Commented by ajfour last updated on 10/Jun/17

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

t g ∠ N B T = t g ∠ A D M = \frac{b}{2 a} \Rightarrow Q M ∥ P N (1)

t g ∠ Q C D = t g ∠ T A B = \frac{a}{2 b} \Rightarrow P Q ∥ M N (2)

f r o m (1), (2) \Rightarrow M N P Q, i s a p a r a l l e l o g r a m .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Jun/17