Question #41bbd

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Answer 1 · 2017-03-03T09:36:19

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

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