Question #41bbd

1 Answer
Mar 3, 2017

drawn

ABCD is a rhombus. Its diagonals AC and BD intersect at O.

Given #AC+BD =28#
#=>2AO+2BO=28#
#=>AO+BO=14#

Again #m/_CBA=2m/_DAB#

Let #m/_BAO=m/_BCO=x^@#

So #m/_DAB=2x^@#

Hence #m/_CBA=4x^@#

Considering the sum of three angles of #DeltaABC#

we get #x+4x+x=180^@#

#=>x=30^@#

In #Delta AOB,#

# /_AOB=pi/2" (diagonals of rhombus bisect vertically)" #

So #AO=ABcos /_BAO=ABcos30=sqrt3/2AB#

#BO=ABcos /_BAO=ABsin30=1/2AB#
Again
#=>AO+BO=14#

So #=>sqrt3/2AB+1/2AB=14#

#=>AB=28/(sqrt3+1)=(28(sqrt3-1))/2=14(sqrt3-1)#

Hence the perimeter of the rhombus

#=4AB\=56(sqrt3+1)#