Question-209314

Question Number 209314 by SonGoku last updated on 06/Jul/24

Commented by SonGoku last updated on 06/Jul/24

How to find the of this polygon ?

Commented by mr W last updated on 07/Jul/24

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

t h e p o l y g o n, n o t h i n g e l s e o f i t .

Commented by SonGoku last updated on 07/Jul/24

So the only way to determine the diagonal of ang

irreular polygon, like the one in the image, is onlyo

thrugh practice ? In other words, in the field ?

Commented by Frix last updated on 07/Jul/24

Diagonal = d

18 < d < 35

\frac{\sqrt{197319}}{4} < area ⩽ \frac{\sqrt{1278519}}{4}

(\min at d = 35, \max at d = \frac{\sqrt{638290}}{29})

Commented by Frix last updated on 07/Jul/24

For 1 triangle you need at least 3 measurements .

In this case, you need 1 additional measurment

for one of the triangles, the rest follows .

Commented by SonGoku last updated on 08/Jul/24

But, how did you at these calculations ?

Commented by Frix last updated on 08/Jul/24

1^{st} triangle 10, 28, d \Rightarrow 18 < d < 38

2^{nd} triangle 15, 20, d \Rightarrow 5 < d < 35

\Rightarrow 18 < d < 35

Area of triangle a, b, c

= \frac{\sqrt{(a + b + c) (b + c - a) (a + c - b) (a + b - c)}}{4}

Area of 1^{st} triangle

A_{1} = \frac{\sqrt{- d^{4} + 1768 d^{2} - 467856}}{4}

Area of 2^{nd} triangle

A_{2} = \frac{\sqrt{- d^{4} + 1250 d^{2} - 30625}}{4}

For the minimum obviously d = 18 \Rightarrow A_{1} = 0

For the maximum I used differentiation

Commented by mr W last updated on 08/Jul/24

m a x i m u m a r e a i s w h e n t h e

q u a d i l a t e r a l i s c y c l i c .

s = \frac{a + b + c + d}{2} = \frac{10 + 28 + 15 + 20}{2} = 36.5

A_{m a x} = \sqrt{(s - a) (s - b) (s - c) (s - d)}

= \sqrt{(36.5 - 10) (36.5 - 28) (36.5 - 15) (36.5 - 20)}

= \frac{\sqrt{1278519}}{4}

s e e a l s o Q 30233

Commented by SonGoku last updated on 09/Jul/24

Very sophisticated .

Congratulations

I willa study .

Thank you .

Commented by SonGoku last updated on 09/Jul/24

So, can I use this formula to determine the area of

any quadrilateral ?

Commented by mr W last updated on 09/Jul/24

n o! {i t}^{'} s o n l y f o r s o - c a l l e d c y c l i c

q u a d r i l a t e r a l s .

g e n e r a l l y a q u a d r i l a t e r a l i s n o t

u n i q u e l y d e f i n e d w h e n o n l y i t s f o u r

s i d e s a r e g i v e n . y o u n e e d a n

a d d i t i o n a l c o n d i t i o n .

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