Lesson Area of a quadrilateral

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Area of a quadrilateral


Theorem 1

The area of a convex quadrilateral equals half the product of the measures of its            
diagonals d%5B1%5D  and  d%5B2%5D  and the sines of the angle between them.

S = 1%2F2.d%5B1%5D.d%5B2%5D.sin%28delta%29            (1)



  Figure 1a. To the Theorem 1

Proof

The diagonals  AC  and  BD  divide the quadrilateral  ABCD  in four triangles  ABP,  BCP,  CDP  and  ADP,  where  P  is the intersection point of the diagonals  (Figure 1b).

For each of these four tringles, the area is equal to                                                            

S%5BADP%5D = %281%2F2%29|AP|.|DP|.sin%28delta%29,
S%5BABP%5D = %281%2F2%29|AP|.|BP|.sin%28pi-delta%29 = %281%2F2%29|AP|.|BP|.sin%28delta%29,
S%5BBCP%5D = %281%2F2%29|BP|.|CP|.sin%28delta%29,
S%5BDCP%5D = %281%2F2%29|DP|.|CP|.sin%28pi-delta%29 = %281%2F2%29|DP|.|CP|.sin%28delta%29



Figure 1b. To the proof of the Theorem 1

in accordance with the formula  (3)  for the area of a triangle of the lesson  Formulas for area of a triangle  in this site.

The area of the quadrilateral is the sum of the areas of the four triangles
S = S%5BADP%5D + S%5BABP%5D + S%5BBCP%5D + S%5BDCP%5D = 1%2F2( |AP|.|DP|.sin%28delta%29 + |AP|.|BP|.sin%28delta%29 + |BP|.|CP|.sin%28delta%29 + |DP|.|CP|.sin%28delta%29 ).

Let us transform the right side step by step to get the required result.

S = 1%2F2( |AP|.|DP| + |AP|.|BP| ).sin%28delta%29 + 1%2F2( |BP|.|CP| + |DP|.|CP| ).sin%28delta%29 = 1%2F2.|AP|.( |DP| + |BP| ).sin%28delta%29 + 1%2F2.|CP|.( |BP| + |DP| ).sin%28delta%29 =
1%2F2.|AP|.d%5B2%5D.sin%28delta%29 + 1%2F2.|CP|.d%5B2%5D.sin%28delta%29 = 1%2F2.( |AP| + |CP| ).d%5B2%5D.sin%28delta%29 = 1%2F2.d%5B1%5D.d%5B2%5D.sin%28delta%29.

It is what has to be proved.


Example 1

Find the area of a quadrilateral if its diagonals are of  18 cm  and  12 cm  long and the angle between the diagonals is of 50°.

Solution

Apply the formula  (1)  above.  According to this formula,  the area of the quadrilateral is equal to  1%2F2*18*12*sin(50°) = 0.5*18*12*0.766 = 82.728 cm%5E2  (approximately).

Answer.  The area of the quadrilateral is  82.728 cm%5E2  (approximately).


My other lessons on the topic  Area  in this site are
    - WHAT IS area?
    - Formulas for area of a triangle
    - Proof of the Heron's formula for the area of a triangle
    - One more proof of the Heron's formula for the area of a triangle
    - Proof of the formula for the area of a triangle via the radius of the inscribed circle
    - Proof of the formula for the radius of the circumscribed circle
    - Area of a parallelogram
    - Area of a trapezoid
    - Area of a quadrilateral circumscribed about a circle
    - Area of a quadrilateral inscribed in a circle
under the topic  Area and surface area  of the section  Geometry,  and
    - Solved problems on area of triangles
    - Solved problems on area of right-angled triangles
    - Solved problems on area of regular triangles
    - Solved problems on the radius of inscribed circles and semicircles
    - Solved problems on the radius of a circumscribed circle
    - A Math circle level problem on area of a triangle
    - Solved problems on area of parallelograms
    - Solved problems on area of rhombis, rectangles and squares
    - Solved problems on area of trapezoids
    - Solved problems on area of quadrilaterals
under the topic  Geometry  of the section  Word problems.

For navigation over the lessons on  Area of Quadrilaterals  use this file/link  OVERVIEW of lessons on area of quadrilaterals.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

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