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Prove that the area of a quadrilateral is one half the product of the lengths of its diagonals and the sine of the angle between the diagonals.
enter image description here

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    $\begingroup$ What are your thoughts? (Hint: Decompose the quadrilateral into two triangles). $\endgroup$ Mar 16, 2013 at 15:36
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    $\begingroup$ Nitpicking: if your angle $\theta$ is oriented, it should be $|\sin\theta|$ in the formula. $\endgroup$
    – Julien
    Mar 16, 2013 at 15:39
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    $\begingroup$ As I know it can be proved by drawing parallel to diagonals lines through the points of the quadrilateral. $\endgroup$ Mar 2, 2018 at 22:40

3 Answers 3

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Hint:

Let $S$ be the area of a triangle, $a,b$ be the length of two edges and $\theta$ be the angle between them, you have the following formula $$S=\frac{1}{2}ab\sin\theta$$

On more hint: $\sin\theta=\sin(\pi-\theta)$.

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  • $\begingroup$ I was also thinking about these formulas but how to use them? $\endgroup$
    – Reader
    Mar 16, 2013 at 15:43
  • $\begingroup$ Try to expand $(a+b)(c+d)$ $\endgroup$
    – NECing
    Mar 16, 2013 at 15:47
  • $\begingroup$ I have done that. I have four triangles in the figure above. and the area I get is 1/2(a+b)(c+d) but the problem is where do I get sine from? $\endgroup$
    – Reader
    Mar 16, 2013 at 16:00
  • $\begingroup$ Look at the formula again. $\sin\theta$ is in it. $\endgroup$
    – NECing
    Mar 16, 2013 at 16:02
  • $\begingroup$ I am confused now , I have the proof but still I dont have. $\endgroup$
    – Reader
    Mar 16, 2013 at 16:04
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The area of this quadrilateral can be calculated by summing up the area of its four triangles. For a triangle, its area can be calculated using the formula:
$A=\frac{1}{2}ab\sin\theta$
where $a$ and $b$ are the lengths of two of his sides and $\theta$ is the internal angle between them, so the total area of the quadrilateral is:
$A=\frac{1}{2}ac\sin\theta_1+\frac{1}{2}cb\sin\theta_2+\frac{1}{2}bd\sin\theta_3+\frac{1}{2}da\sin\theta_4$
where the four angles $\theta_1$, $\theta_2$, $\theta_3$ and $\theta_4$ are the internal angles between each couple of edges. By knowing that $\theta_1$ = $\theta_3$ and $\theta_2$ = $\theta_4$, and $\theta_1$ and $\theta_3$ are supplementary angles ($\theta_1 = \pi-\theta_3$), so all the angles have the same $sin$, hence:
$A=\frac{1}{2}ac\sin\theta+\frac{1}{2}cb\sin\theta+\frac{1}{2}bd\sin\theta+\frac{1}{2}da\sin\theta\\=\frac{1}{2}(ac+cb+bd+da)\sin\theta\\=\frac{1}{2}(a+b)(c+d)\sin\theta\\=\frac{1}{2}d1\cdot d2\sin\theta$
where $d1$ and $d2$ are the lengths of the two diagonlas and $\theta$ is any internal angle between them.

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    $\begingroup$ +1. GREAT answer! I noticed you "followed" my Materials Modeling stack exchange site proposal but did not yet click "commit". Would you be able to please commit to it, as we are getting very desperate now to reach 100 committers with 200 rep? We already reached allother requirements! area51.stackexchange.com/proposals/122958/…. It would be very much appreciated! Thank you! $\endgroup$ Apr 4, 2020 at 19:59
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!**enter image description here ** **The proof:

Area of a quadrilateral = area of the first triangle + area of the second triangle + area of the third triangle + area of the fourth triangle

= [0.5 .d . b . sin θ ] + [0.5 . b . c . sin (180 - θ) ] + [0.5 . a . c . sin θ ] + [0.5 . d . a . sin (180 - θ) ]

As, sin (180 - θ) = sin θ

Area = [0.5 . d . b . sin θ ] + [0.5 . b . c . sin θ ] + [0.5 . a . c . sin θ ] + [0.5 . d . a . sin θ ]

Area = 0.5 . sin θ . [ (d . b) + (b . c) + (a . c) + (d . a) ]

As, (d . b) + (b . c) = b . (c + d)
and (a . c) + (d . a) = a . (c + d)

Area = 0.5 . sin θ . [ b . (c + d) + a . ( c + d) ]

Area = 0.5 . sin θ . (c + d) . [b + a]

As, (b + a) = p , (c + d) = q

Area = 0.5 . p . q . sin θ

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    $\begingroup$ Welcome to mathstack exchange. In the future, please use MathJax in order to make your answer more readable $\endgroup$
    – QC_QAOA
    Dec 26, 2019 at 18:41
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    $\begingroup$ What are $diag1$ and $diag2$? $\endgroup$
    – an4s
    Dec 26, 2019 at 19:22

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