Show that for any parallelogram,if a and b are the sides of two non parallel sides,x is the angle between these 2 sides and d is the length of the diagonal that has common vertex with sides a and b,then d^2=a^2+b^2+2ab(cosx).

Asked by  | 1st Mar, 2013, 11:33: AM

Expert Answer:

Answer Given : for any parallelogram,if a and b are the sides of two non parallel sides,x is the angle between these 2 sides and d is the length of the diagonal that has common vertex with sides a and b
To prove : d2=a2+b2+2ab(cosx)
 
Let ABCDbe the given parallelogram and 
AB =a
AD=b
AC =d
angle BAD = x
 
Since ABCD is an llgm , therefore
BC=AD= b { as opposite sides of llgm are equla}
 
also
in an llgm sum of all angles is 360 degree or 2pi
=> angle ABC + angle ADC = 2pi - 2x  { angle DAB = angle angle BCD = x as opposite angles in llgm are equla }
 
=>2 * angle ABC = 2pi-2x   { angle ABC = ange ADc as opposite sides  of llgm are equal}
 
=> angle ABC = (2pi -2x) /2
=> angle ABC = pi -x 
 
now in triangle ABC 
cos( angle ABC) = cos (pi-x) = - cos x
 
By cosine law
cos(angle ABC) = - cosx = (a2 +b2 -d2) /2ab 
                    =>  a2 +b2 + 2ab cos x = d2
Hence proved

Answered by  | 1st Mar, 2013, 03:59: PM

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