Question

Each side of rhombus is 10 centimetre long and one of its diagonal measures 16 cm find the length of the Other diagonal .​

Answer 1

Let us consider ABCD is a rhombus and AC and BD are the diagonal and O is the point of intersection of diagonal.

Given :-

Each sides of rhombus be 10 cm

AB = BC = CD = AD = 10 cm

Length of one diagonal is 16 cm ( Let AC be 16 cm )

To Find :-

Length of other diagonal which is BD.

Solution :-

In rhombus ABCD,

AO = OC &

BO = OD [ Diagonals of rhombus bisect each other at 90° ]

Therefore, AO = OC = 8 cm

$\mathrm{In\:\triangle\: AOB,}$

AB = 10 cm

Angle AOB = 90°

AO = 8 cm

Therefore, by Pythagoras theorem

AB^2 = AO^2 + BO^2

10^2 = 8^2 + BO^2

100 = 64 + BO^2

BO^2 = 100 - 64

BO^2 = 36

BO = 6 cm

Since = BO = OD = 6 cm

Therefore, BD = 12 cm

Answer 2

According to the given question:-

AB = BC = CD = AD = 10 cm

AO = OC

BO = OD

AO= OC = 8 cm

Now, Triangle AOB

AB = 10 cm

AOB = 90°

AO = 8 cm

Using Pythagoras theorem:-

AB^2 = AO^2 + BO^2

10^2 = 8^2 + BO^2

100 = 64 + BO^2

BO^2 = 100 - 64

BO^2 = 36

BO = 6 cm

Therefore,= BO = OD = 6 cm

Hence, BD = 12 cm