Question

The lengths of the diagnols of a rhombus are16 and 12 cm find the side of the rhombus

Answer 1

⇒ Let ABCD be the rhombus and AC and BD bisect at point O. AC = 16cm and BD = 12cm.

⇒ We know that the diagonals of rhombus bisect at right angles.

⇒ AO=

2

16

=8cm

⇒ BO=

2

12

=6cm

⇒ In right angled △AOB,

By using Pythagoras theorem,

⇒ AB

2

=AO

2

+BO

2

⇒ AB

2

=8

2

+6

2

⇒ AB

2

=64+36

⇒ AB

2

=100

⇒ AB=

100

⇒ AB=10cm

∴ Side of a rhombus is 10cm.

solution

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Answer 2

As the question mentioned there are 2 diagonals

The length of the first diagonal is 12 cm, whereas the length of the second diagonal is 16 cm

We divide the rhombus into 4 parts, let us take part 1:

As the diagram shown below, using Pythagoras theorem we find s

Now as we know that BD = 16 cm and AC = 12 cm; therefore to find S

We use,

${(bd \div 2)}^{2} + {(ac \div 2)}^{2} = {s}^{2}$

Putting the values of BD and AC, we get,

${(16 \div 2)}^{2} + {(12 \div 2)}^{2} = {s}^{2}$

${8}^{2} + {6}^{2} = {s}^{2}$

$\sqrt{100} = {s}^{2}$

s=10cm