Archit3110 wrote:
Bunuel wrote:
What is the perimeter of a rhombus?
(1) Its area is 140 square meters
(2) One of its diagonals is 48 meters
diagnoal of rhombus bisect at 90* and adjacents sides are equal so if we know diagonal length we can find perimeter
from 1
area of rhombus : 0.5 * d1*d2
d1*d2= 280
d1 and d2 not know
in sufficient
from 2
d1 = 48
so two sides would be
2x^2= 48
x^2=24
x= 2 sqrt6
we dont know other two sides so in sufficient
from 1 & 2
d2* 48= 280
d1= 5.8 ~ 6
so other two sides
2x^2= 6
x^2= 3
x= sqrt 3 = ~1.7 = ~ 2
so perimeter = 2 * ( 2+ 2 sqrt6 )= 2* ~7 = 14
IMO C
Hi
Archit3110,
Thank you for your solution But I am unable to get the following part,May be I am missing something,
Archit3110 wrote:
d1 = 48
so two sides would be
2x^2= 48
x^2=24
x= 2 sqrt6
How do we know that the two adjacent sides form a 90 Deg. angle to use Pythagoras theorem. I assume you meant to write \(2x^2= 48^2\), where x is one side of a rhombus. Also if we knew X, shouldn't we know all the sides of the rhombus as in a Rhombus all the sides are equal.
Thank you.
area of rhombus = 1/2 * d1 * d2 ; d1 & d2 are two diagonal of rhombus which bisect each other at 90*..