# Area of a quadrilateral proof

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.

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

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: $\sin\theta=\sin(\pi-\theta)$.

• I was also thinking about these formulas but how to use them? Mar 16, 2013 at 15:43
• Try to expand $(a+b)(c+d)$ Mar 16, 2013 at 15:47
• 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? Mar 16, 2013 at 16:00
• Look at the formula again. $\sin\theta$ is in it. Mar 16, 2013 at 16:02
• I am confused now , I have the proof but still I dont have. Mar 16, 2013 at 16:04

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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!**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

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