Each side of a rhombus is 5cm,one of the diagonals is 8cm.Calculate length of other diagonal,area of rhombus.?

Answer 1

Let rhombus be ABCD with diagonals that intersect at E.
AB = 5, AE = 4 , EB = x , say
5² = 4² + x²
x² = 9
x = 3
Diagonal DB = 6 cm
Area = (1/2) x 6 x 8 cm ²
Area = 24 cm ²

Answer 2

Well if each side is the same you're just finding a triangle and multiplying it. You know that the base is 24/2 on the triangle and the hypoteneuse is 13 riiight? so.... 12^2 + b^2 = 13^2 b^2 = 13^2 - 12^2 b^2 = 169 - 144 b^2 = 25 b = 5 now since we have only solved one half of the rhombus with 5 we simply double it and we find that the other is - 10 cm

Answer 3

It's 6cm
If you take half the length of your diagonal (4cms) it makes a rightangle triangle with the side of your rhombus and half of the other diagonal. The side of your rhombus is the longest side, which I think is the hypotenuse (but don't quote me on that) To work out the length of one side of a rightangle triangle the equation is a(squared)+b(squared)=c(squared) where c is the hypotenuse and a and b are the other two sides.
Therefore you know the hypotenuse = 5 and one of the other sides = 4 so you should change the equation from
a(squared)+b(squared)=c(squared) to
c(squared)-a(squared)=b(squared) thus
5 squared - 4 squared = b squared
thus b squared= 25-16 = 9
the square root of 9 is 3
you want the whole length of the diagonal so 3+3=6 and there is your answer

Answer 4

side=5cm
diagonal1=8cm
diagonals of rhombus bisect each other at 90 degree.
now by applying pythagoras theorem,
other diag. will come out to 6 cm
area of rhombus=2[1/2(8[3]
2[1/2[24]
=24cm^2

Answer 5

5^2= 4^2+ x^2then x =3then length of other diagonal2x3=6 cm
area 1/2 x 8 x 6 sin 90= 24 cm^2