Question

find the length of the second diagonal of a rhombus whose side is 5cm and one of the diagonals is 8cm long.​

Answer 1

Answer:

Since we know that the diagonals of a rhombus are perpendicular bisectors of each other, lets assume that ABCD is a rhombus and it's diagonals intersect at point O. Since all the sides of the rhombus are equal therefore AB=BC=CD=DA=5cm and let say diagonal AC=8cm. Since the diagonals intersect at point O this implies O is the mid-point of the diagonals AC & BD i.e. AO=4cm and OC=4cm. Also the diagonals of the rhombus bisect perpendicularly this implies that angle AOB = 90°I.e. ΔAOB is a right angled triangle with AO=3cm,AB=5cm and OB=x cm"OD=x hence,OB=OD then by Pythagoras theorem we can write,

$oa ^{2} + ob ^{2} = ab ^{2}$

${4}^{2} + {x}^{2} = {5}^{2}$

$16 + {x}^{2} = 25$

${x}^{2} = 25 - 16$

${x}^{2} = 9$

$x = \sqrt{9}$

$x = + 3andx = - 3$

But since x is lenght it cannot be negative hence x=3cm.

OB =x And x=3

we know that OB=OD,

So,OD=3cm

We have to find BD,

Hence,

OB+OD=3+3

=6cm

Hence 6cm is the second diagonal length

Hope it helps

Answer 2

Answer:

Answer is 6. Mark as brilliant