Question

Question #41bbd

Answer 1

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)`