Question

find the perimeter of a rhombus with diagonals that measure 30 and 40

Answer 1

\[\sqrt{15^{2}+20^{2}}=25\]
\[25*4=100\]

Answer 2

the diagonals of a rhombus parts the rhombus in 4 triangles equals (the same) and we know again that the diagonals in a rhombus divides in half parts (1/2) - in case of a rhombus ABCD in this way we can us to calculate the lenghst of AB or BC or CD or AD - Pythagoras's theorem - if AB=a,BC=b,CD=c and DC=d and the point section of diagonals being O and hence for one side AB we can writing (AB)*2=(AO)*2+(OB)*2 what will be a*2=(15)*2 +(20)*2 ---a*2=225+400=625 --- a=25 but because a=b=c=d resulted that the perimeter equal a+b+c+d=25+25+25+25=100