Question

the area of rhombus is 120cm2 and one of its diagonal is 24cm. find each side of rhombus

Answer 1

120=1/2*24*#
#=10
consider rhombus and
then,
ad^2=5^2+12^2=25+144=169 √169=13=side of rhombus

Answer 2

See brother...,
Rhombus has diagonals intersecting at 90°,
So, Let ABCD be a rhombus and O be the point of intersection of the diagonals.....
Then... First of all...
Area of rhombus=120cm^2
So,
1/2×product of its diagonals=120cm^2
So,
1/2×24cm×2nd diagonal=120cm^2
Then,
2nd diagonal=2×120/24=10cm
So here begins the solution...
Now,
AO=1/2 of AC=1/2×24cm=12cm... (1)
And,
BO=1/2 of BD=1/2×10cm=5cm.... (2)
Now In triangle AOB.... By Pythogoras theorem,
AB^2=AO^2+BO^2
AB=√12^2+5^2
AB=√144+25
AB=√169=13
So eac side of rhombus is 13cm....
Please like if u get it.....