prove that the area of a convex quadrilateral is 1/2 the product of the lenghts of the diagonals?

Question

prove that the area of a convex quadrilateral is 1/2 the product of the lenghts of the diagonals and the sine of the angle between the diagonals.

[ please answer, i need it tomorrow ]

Answer 1

Please follow the alphabet naming and you will get it(since I cant show you the figure in Y!A)

Let the convex quadrilateral be ABCD

Draw one Diagonal say BD

You get two triangles ABD & triangle BCD

Drop a perpendicular segment from vertex A on diagonal BD to get AE

Drop another perpendicular segment from vertex C on diagonal BD to get CF

These perpendiculars represent the altitude(height) of Tri(ABD) and Tri(BCD) respectively.

Area of Triangle ABD = ½*base*height = ½*BD*AE

Area of Triangle BCD = ½*base*height = ½*BD*CF

Area of quadrilateral ABCD = A(Triangle ABD) + A(Triangle BCD)

= ½*BD*AE + ½*BD*CF

= ½*BD(AE + CF) ........... (I)

Angle between diagonals AC and BD is say ¢ º
Let diagonals AC and BD intersect at a point O

So from given data angle AOE = angle COF = ¢ º

In small triangle AEO:
AE = AO*sin(¢)

In small triangle COF:
CF = CO*sin(¢)

Substitute values of AE & CF in relation ......... (I)

Area of quadrilateral ABCD

= ½*BD*[AO*sin(¢) + CO*sin(¢)]

= ½*BD*sin(¢)[AO + CO]

= ½*BD*AC*sin(¢) .............. since AC = AO + OC

THUS, the area of a convex quadrilateral is 1/2 the product of the lenghts of the diagonals and the sine of the angle between the diagonals.

Answer 2

Area Of Convex Quadrilateral